A centrifuge has an angular velocity of 3,000 rpm, what is the acceleration (in unit of the earth gravity) at a point with a radius of 10 cm

Answer:

Wavelength is 0.359 m

Explanation:

Given that,

Magnetic field, B = 0.547 T

We need to find the wavelength of radiation produced by a proton in a cyclotron with a magnetic field of 0.547 T.

The frequency of revolution of proton in the cyclotron is given by :

[tex]f=\dfrac{qB}{2\pi m}[/tex]

m is mass of proton

q is charge on proton

So,

[tex]f=\dfrac{1.6\times 10^{-19}\times 0.547}{2\pi \times 1.67\times 10^{-27}}\\\\f=8.34\times 10^6\ Hz[/tex]

We know that,

Speed of light, [tex]c=f\lambda[/tex]

[tex]\lambda[/tex] = wavelength

[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{834\times 10^6}\\\\\lambda=0.359\ m[/tex]

So, the wavelength of the radiation produced by a proton is 0.359 m.

Answer:

The horizontal force acting on m2 is F + 9.8m1

Explanation:

Given;

Block m1 on left of block m2

Make a sketch of this problem;

                         F →→→→→→→→→→→-------m1--------m2

Apply Newton's second law of motion;

F = ma

where;

m is the total mass of the body

a is the acceleration of the body

The horizontal force acting on block m2 is the force applied to block m1 and force due to weight of block m1

F₂ = F + W1

F₂ = F + m1g

F₂ = F + 9.8m1

Therefore, the horizontal force acting on m2 is F + 9.8m1

The force acting on the block of mass m₂ is  [tex]\frac{m_2F}{m_1+m_2}[/tex]

Given that there are two blocks of mass m₁ and m₂.

m₁ is on the left of block m₂. They are in contact with each other.

A force F is applied on m₁ to the right.

According to Newton's laws of motion:

The equation of motion of the blocks can be written as:

F = (m₁ + m₂)a

here, a is the acceleration.

so, acceleration:

a = F / (m₁ + m₂)

Now, the force acting on the block of mass m₂ is:

f = m₂a

[tex]f = \frac{m_2F}{m_1+m_2}[/tex]

Learn more about laws of motion:

https://brainly.com/question/26083484?referrer=searchResults

Answer:

λ = 1.4 × 10^(-7) m

Explanation:

We are given;

distance of eye piece from the source;D = 1.5 m

distance between the virtual sources;d = 7.5 × 10^(-4) m

To find the wavelength, we will use the formula for fringe width;

X = λD/d

Where X is fringe width, λ is wavelength, while d and D remain as before.

Now, fringe width = eye-piece distance moved transversely/number of fringes

Eye piece distance moved transversely = 1.88 cm = 1.88 × 10^(-2) m

Thus,

Fringe width = (1.88 × 10^(-2))/10 = 1.88 × 10^(-3) m

Thus;

1.88 × 10^(-3) = λ(1.5)/(7.5 × 10^(-4))

λ = [1.88 × 10^(-3) × (7.5 × 10^(-4))]/1.5

λ = 1.4 × 10^(-7) m

Answer:

F = umg where u is coefficient of dynamic friction

Explanation:

F = 0.4 x 80 x 9.81 = 313.92 N

Answer:

power dissipated in the target is 100 W

Explanation:

given data

electrons = 1 mev = [tex]10^{6}[/tex] eV

1 eV = 1.6 × [tex]10^{-19}[/tex] J

current =  100 microampere = 100 × [tex]10^{-6}[/tex] A

solution

when energy of beam strike with 1 MeV so energy of electron is

E = e × v   ...................1

e is charge of electron and v is voltage

so put here value and we get voltage

v = 1 ÷ 1.6 × [tex]10^{-19}[/tex]

v =  [tex]10^{6}[/tex] volt

so power dissipated in target

P = voltage × current   ..............2

put here value

P =  [tex]10^{6}[/tex]  × 100 × [tex]10^{-6}[/tex]

P = 100 W

so power dissipated in the target is 100 W

Answer:

  F ’= F 0.25

Explanation:

This problem refers to the electric force, which is described by Coulomb's law

        F = k q₁ q₂ / r²

where k is the Coulomb constant, q the charges and r the separation between them.

The initial conditions are

       F = k q_A q_B / d²

they indicate that the loads are reduced to ¼ q and the distance is reduced to ½ d

       F ’= k (q / 4 q / 4) / (0.5 d)²

       F ’= k q / 16 / 0.25 d²

       F ’= k q² / d²    0.0625 / 0.25

       F ’= F 0.25

Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. If the charges are reduced to one-fourth of their original values and the distance is halved, the new force will be one-fourth of the original force.

Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. This can be explained through Coulomb's law.

Coulomb's law is a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.

[tex]F = k \frac{q_Aq_B}{d^{2} } = k \frac{q^{2} }{d^{2} } [/tex]

where,

If you reduce the charge of A to one-fourth its original value, and the charge of B to one-fourth, and reduce the distance between the objects by half, the new force will be:

[tex]F_2 = k \frac{(0.25q_A)(0.25q_B)}{(0.5d)^{2} } = 0.25k\frac{q^{2} }{d^{2} } = 0.25 F[/tex]

Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. If the charges are reduced to one-fourth of their original values and the distance is halved, the new force will be one-fourth of the original force.

Learn more about Coulomb's law here: https://brainly.com/question/506926

Answer:

A. 1.6 N/cm

Explanation:

spring constant = 21/13 = 1.6 N/cm

Answer:

The  length is  [tex]D = 5 \ cm[/tex]

Explanation:

From the question we are told  that

     The  length of the  hand is  [tex]l = 10.0 \ cm[/tex]

      The  angle at the hand is  held is  [tex]\theta = 30 ^o[/tex]

Generally resolving the length the length of the hand to it vertical component we obtain that the length of the shadow on the vertical wall is mathematically evaluated as

             [tex]D = l * sin(\theta )[/tex]

substituting values

             [tex]D = 10 * sin (30)[/tex]

             [tex]D = 5 \ cm[/tex]

Answer:

U =-2.39*10^-18 J

Explanation:

In order to calculate the electric potential energy of the electron you use the following formula:

[tex]U=k\frac{q_1q_2}{r}[/tex]           (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

r: distance between charges

In this case the electron is at point midway between two charges, then the electric potential energy is the sum of two contributions:

[tex]U=U_1+U_2=k\frac{eq_1}{r}+k\frac{eq_2}{r}=\frac{ke}{r}[q_1+q_2][/tex]

e: charge of the electron = 1.6*10^-19C

q1: charge 1 = 3.00nC = 3.00*10^-9C

q2: charge 2 = 2.00nC = 3.00*10^-9C

r: distance to each charge = 60.0cm/2 = 30.0cm = 0.3m

If you consider that the electron is at the origin of coordinates, with the first charge in the negative x axis, and the other one in the positive x axis, you have:

[tex]U=\frac{(8.98*10^9Nm^2/C^2)(1.6*10^{-19}C)}{0.6m}[-3.0*10^{-9}C+2.0*10^{-9}C]\\\\U=-2.39*10^{-18}J[/tex]

The electric potential energy of the electron is -2.39*10^-18 J

Answer:

(a) force is 1066.56N

Explanation:

(a) MV²/R

Answer:

The final temperature of the metals will be 384.97 K

Explanation:

For the gold;

mass = 1.3 kg

temperature = 300 K

specific heat =  126 J/kg-K

For the copper;

mass = 2.4 kg

temperature = 400 K

specific heat = 386 J/kg-K

Firstly, we will have to calculate for the thermal energy possessed by each of the metal.

The heat possessed by a body =  mcT

Where,

m is the mass of the body

c is the specific heat of the body, and

T is the temperature of the body at that instance

so we calculate for the thermal energy of the gold and the copper below

For gold;

heat energy = mcT = 1.3 x 126 x 300 = 49140 J

For copper;

heat energy = mcT = 2.4 x 386 x 400 = 370560 J

When the two metal come in thermal contact, this heat is evenly distributed between them.

The total heat energy = 49140 J + 370560 J = 419700 J

At thermal equilibrium, the two metals will be at the same temperature, to get this temperature, we equate the total thermal energy to the heat energy that will be possessed by the metals at equilibrium.

419700 = (1.3 x 126 x T) + (2.4 x 386 x T) = 163.8T + 926.4T

419700 = 1090.2T

T = 419700/1090.2 = 384.97 K

The final temperature of the metals will be 384.97 K

The period of a wave is the time it takes the wave to complete one cycle (at a fixed location).

So if a wave completes one cycle in 2 seconds, then that is its period.

Answer:

u = 10.02m/s

Explanation:

a = f/m

a = 20/12 = 1.67m/s²

U =2aS

u = 2 x 1.67 x 3

U = 10.02m/s

Answer:

The resistivity is  [tex]\rho = 2.5 *10^{-8} \ \Omega \cdot m[/tex]

Explanation:

From the question we are told that

    The magnitude of the electric field is  [tex]E = 6.2 V/m[/tex]

     The  current density is  [tex]J = 2.4 *10^{8} \ A/m^2[/tex]

Generally  the resistivity is mathematically represented as

         [tex]\rho = \frac{E}{J}[/tex]

substituting values

        [tex]\rho = \frac{6.2}{2.4 *10^{8}}[/tex]

        [tex]\rho = 2.5 *10^{-8} \ \Omega \cdot m[/tex]

Answer:

Choice d. The set of vectors: [tex]\displaystyle \left\{\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix},\; a,\, b\in \mathbb{R} \right\}[/tex] isn't a vector space over [tex]\mathbb{R}[/tex].

Explanation:

Let a set of vectors [tex]V[/tex] to be a vector field over some field [tex]\mathbb{F}[/tex] (for this question, that "field" is the set of all real number.) The following must be true:

Note that in the general form of a vector in [tex]V[/tex], the second component is a always non-zero. Because of that non-zero component,

Assume by contradiction that [tex]V[/tex] is indeed a vector field. Therefore, it should contain a zero vector. Let [tex]\mathbf{0}[/tex] denote that zero vector. For all [tex]\mathbf{v} \in V[/tex], [tex]\mathbf{v} + \mathbf{0} = \mathbf{v}[/tex].

Using the definition of set [tex]V[/tex]: [tex]\displaystyle \left\{\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix},\; a,\, b\in \mathbb{R} \right\}[/tex], there exist real numbers [tex]a[/tex] and [tex]b[/tex], such that:

[tex]\displaystyle \mathbf{v} = \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix}[/tex].

Hence, [tex]\mathbf{v} + \mathbf{0} = \mathbf{v}[/tex] is equivalent to:

[tex]\displaystyle \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} + \mathbf{0} = \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix}[/tex].

Apply the third property that [tex]V[/tex] is closed under scalar multiplication. [tex]-1[/tex] is indeed a real number. Therefore, if [tex]\mathbf{v}[/tex] is in

Therefore:

[tex]\displaystyle -\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} \in V[/tex].

Apply the second property and add [tex]\displaystyle - \mathbf{v} = -\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix}[/tex] to both sides of [tex]\mathbf{v} + \mathbf{0} = \mathbf{v}[/tex]. The left-hand side becomes:

[tex]\mathbf{v} - \mathbf{v} + \mathbf{0} = \mathbf{0}[/tex].

The right-hand side becomes:

[tex]\displaystyle \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} -\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} = \begin{bmatrix}a - 4\, b - (a - 4\, b) \\ 5 - 5 \\ 4\, a+ b-(4\, a+ b)\\ -a -b - (-a -b)\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0 \\0\end{bmatrix}[/tex].

Therefore:

[tex]\displaystyle \mathbf{0} = \begin{bmatrix}0 \\ 0 \\ 0 \\0\end{bmatrix}[/tex].

However, [tex]\mathbf{0} = \displaystyle \begin{bmatrix}0 \\ 0 \\ 0 \\0\end{bmatrix}[/tex] isn't a member of the set [tex]\displaystyle V = \left\{\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix},\; a,\, b\in \mathbb{R} \right\}[/tex]. That's a contradiction, because [tex]\mathbf{0}[/tex] was supposed to be part of [tex]V[/tex].

Hence, [tex]V[/tex] isn't a vector space by contradiction.

Answer:

did you mean released?

Explanation:

If so the process is called respiration

Answer:

Explanation:

check it out and rate me

Answer:

A.  Z = 185.87Ω

B.  I  =  0.16A

C.  V = 1mV

D.  VL = 68.8V

E.  Ф = 30.59°

Explanation:

A. The impedance of a RL circuit is given by the following formula:

[tex]Z=\sqrt{R^2+\omega^2L^2}[/tex]       (1)

R: resistance of the circuit = 160-Ω

w: angular frequency = 220 rad/s

L: inductance of the circuit = 0.430H

You replace in the equation (1):

[tex]Z=\sqrt{(160\Omega)^2+(220rad/s)^2(0.430H)^2}=185.87\Omega[/tex]

The impedance of the circuit is 185.87Ω

B. The current amplitude is:

[tex]I=\frac{V}{Z}[/tex]                     (2)

V: voltage amplitude = 30.0V

[tex]I=\frac{30.0V}{185.87\Omega}=0.16A[/tex]

The current amplitude is 0.16A

C. The current I is the same for each component of the circuit. Then, the voltage in the resistor is:

[tex]V=\frac{I}{R}=\frac{0.16A}{160\Omega}=1*10^{-3}V=1mV[/tex]            (3)

D. The voltage across the inductor is:

[tex]V_L=L\frac{dI}{dt}=L\frac{d(Icos(\omega t))}{dt}=-LIsin(\omega t)\\\\V_L=-(0.430H)(160\Omega)sin(220 t)=68.8sin(220t)\\\\V_L_{max}=68.8V[/tex]

E. The phase difference is given by:

[tex]\phi=tan^{-1}(\frac{\omega L}{R})=tan^{-1}(\frac{(220rad/s)(0.430H)}{160\Omega})\\\\\phi=30.59\°[/tex]

Answer:

The spring constant is 60,000 N

The total work done on it during the compression is 3 J

Explanation:

Given;

weight of the girl, W = 600 N

compression of the spring, x = 1 cm = 0.01 m

To determine the spring constant, we apply hook's law;

F = kx

where;

F is applied force or weight on the spring

k is the spring constant

x is the compression of the spring

k = F / x

k = 600 / 0.01

k = 60,000 N

The total work done on the spring = elastic potential energy of the spring, U;

U = ¹/₂kx²

U = ¹/₂(60000)(0.01)²

U = 3 J

Thus, the total work done on it during the compression is 3 J

Answer:

(a) V = 65.625 Volts

(b) V = 131.25 Volts

(c) V = 131.25 Volts

Explanation:

Recall that:

1) in a metal sphere the charges distribute uniformly around the surface, and the electric field inside the sphere is zero, and the potential is constant equal to:

[tex]V=k\frac{Q}{R}[/tex]

2) the electric potential outside of a charged metal sphere is the same as that of a charge of the same value located at the sphere's center:

[tex]V=k\frac{Q}{r}[/tex]

where k is the Coulomb constant ( [tex]9\,\,10^9\,\,\frac{N\,m^2}{C^2}[/tex] ), Q is the total charge of the sphere, R is the sphere's radius (0.24 m), and r is the distance at which the potential is calculated measured from the sphere's center.

Then, at a distance of:

(a) 48 cm = 0.48 m, the electric potential is:

[tex]V=k\frac{Q}{r}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.48} =65.625\,\,V[/tex]

(b) 24 cm = 0.24 m, - notice we are exactly at the sphere's surface - the electric potential is:

[tex]V=k\frac{Q}{r}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.24} =131.25\,\,V[/tex]

(c) 12 cm (notice we are inside the sphere, and therefore the potential is constant and the same as we calculated for the sphere's surface:

[tex]V=k\frac{Q}{R}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.24} =131.25\,\,V[/tex]

Answer:

c) a difference in electric potential

Explanation:

my insta: priscillamarquezz

Answer:

a) 43.98 V

b) E = 21.37 MJ

Explanation:

Parameters given:

Length of cable = 180 km = 180000 m

Diameter of cable = 11 cm = 0.11 m

Radius = 0.11 / 2 = 0.055 m

Current, I = 135 A

a) To find the potential drop, we have to find the voltage across the wire:

V = IR

=> V = IρL / A

where R = resistance

L = length of cable

A = cross-sectional area

ρ = resistivity of the copper wire = 1.72 * 10^(-8) Ωm

Therefore:

V = (135 * 1.72 * 10^(-8) * 180000) / (π * 0.055^2)

V = 43.98 V

The potential drop across the cable is 43.98 V

b) Electrical energy is given as:

E = IVt

where t = time taken = 1 hour = 3600 s

Therefore, the energy dissipated per hour is:

E = 135 * 43.98 * 3600

E = 21.37 MJ (mega joules, 10^6)

Answer:

Approximately [tex]0.45\; \rm m \cdot s^{-1}[/tex].

Explanation:

The mechanical energy of an object is the sum of its potential energy and kinetic energy. Consider this question from the energy point of view:

Mechanical energy of the block [tex]0.04\; \rm m[/tex] away from the equilibrium position:

While the block moves back to the equilibrium position, it keeps losing (mechanical) energy due to friction:

[tex]\begin{aligned}& \text{Work done by friction} = (-0.3\; \rm N) \times (0.04 \; \rm m) = -0.012\; \rm J\end{aligned}[/tex].

The opposite ([tex]0.012\; \rm N[/tex]) of that value would be the amount of energy lost to friction. Since there's no other form of energy loss, the mechanical energy of the block at the equilibrium position would be [tex]0.032\; \rm N - 0.012\; \rm N = 0.020\; \rm N[/tex].

The elastic potential energy of the block at the equilibrium position is zero. As a result, all that [tex]0.020\; \rm N[/tex] of mechanical energy would all be in the form of the kinetic energy of that block.

Given that the mass of this block is [tex]0.020\; \rm kg[/tex], calculate its speed:

[tex]\begin{aligned}v &= \sqrt{\frac{2\, \mathrm{KE}}{m}} \\ &= \sqrt{\frac{2 \times 0.020\; \rm J}{0.20\; \rm kg}} \approx 0.45\; \rm m\cdot s^{-1}\end{aligned}[/tex].

Answer:

Option 5:

The 60W bulb has a greater resistance and a lower current than the 100 W bulb.

Explanation:

We have to compare the resistance and current of both bulbs.

Bulb A

Power = 60 W

Voltage = 110 V

Power is given as:

[tex]P = V^2 /R[/tex]

where V= voltage and R = resistance

[tex]=> 60 = 110^2 / R\\\\R = 201.6 \Omega[/tex]

Power is also given as:

P = IV

where I = current

=> 60 = I * 110

I = 60/110 = 0.54 A

Bulb B

Power = 100 W

Voltage = 110 V

To get resistance:

[tex]100 = 110^2 / R\\\\R = 121 \Omega[/tex]

To get current:

100 = I * 110

I = 100 / 110

I = 0.91 A

Therefore, by comparison, the 60W bulb has a greater resistance and a lower current.

Answer:

a) ±7.08×10⁻⁷ C/m²

b) 1.77×10⁻⁷ C

Explanation:

For a conductor,

σ = ±Eε₀,

where σ is the charge density,

E is the electric field,

and ε₀ is the permittivity of space.

a)

σ = ±Eε₀

σ = ±(8×10⁴ N/C) (8.85×10⁻¹² F/m)

σ = ±7.08×10⁻⁷ C/m²

b)

σ = q/A

7.08×10⁻⁷ C/m² = q / (0.5 m)²

q = 1.77×10⁻⁷ C

Answer:

m = 3 kg

The mass m is 3 kg

Explanation:

From the equations of motion;

s = 0.5(u+v)t

Making t thr subject of formula;

t = 2s/(u+v)

t = time taken

s = distance travelled during deceleration = 62.5 m

u = initial speed = 25 m/s

v = final velocity = 0

Substituting the given values;

t = (2×62.5)/(25+0)

t = 5

Since, t = 5 the acceleration during this period is;

acceleration a = ∆v/t = (v-u)/t

a = (25)/5

a = 5 m/s^2

Force F = mass × acceleration

F = ma

Making m the subject of formula;

m = F/a

net force F = 15.0N

Substituting the values

m = 15/5

m = 3 kg

The mass m is 3 kg

Answer:

The atomic number

Explanation:

Answer:

Explanation:

this is the answer to your question

The electron is going with a velocity of 1.25 × 10⁷ m/s when it hits the positive plate.

According to the law of the conservation of mechanical energy, the total mechanical energy is always conserved by an electron. We can say that the sum of potential energy (U) and kinetic energy (K) is always constant.

K + U = E

Given, the distance between the two parallel plates = 1.5 mm

The potential difference between the plates, V = 450V

The charge on an electron, q = [tex]-1.6\times 10^{-19} C[/tex]

The mass of an electron, m = 9.1× 10⁻³¹ Kg

The change in the potential energy of the charge moving through the potential difference of 450V.

ΔU = qΔV = (-1.6× 10⁻¹⁹)(450)  = -7.2 × 10⁻¹⁷J

From the law of the conservation of mechanical energy, we can write:

K + U = E

ΔK + ΔU = 0

ΔK = -ΔU

1/2mv² = -ΔU

v² = -2ΔU/m

[tex]v^2 =\frac{-2\times (-7.2\times 10{-17})}{9.1\times 10^{-31}}[/tex]

[tex]v=\sqrt{1.58\times 10^{14}}[/tex]

v = 1.25 × 10⁷ m/s

Therefore, the electron is going with the speed of 1.25 × 10⁷ m/s  when it hits the positive plate.

Learn more about the law of the conservation of mechanical energy, here:

https://brainly.com/question/20438334

#SPJ5

Answer:

The  corresponding  magnetic field is  

Explanation:

From the question we are told that

    The electric field amplitude is  [tex]E_o = 611\ V/m[/tex]

Generally the  magnetic  field amplitude is  mathematically represented as

              [tex]B_o = \frac{E_o }{c }[/tex]

Where c is the speed of light with a constant value

         [tex]c = 3.0 *0^{8} \ m/s[/tex]

So  

        [tex]B_o = \frac{611 }{3.0*10^{8}}[/tex]

         [tex]B_o = 2.0 4 *10^{-6} \ Vm^{-2} s[/tex]

Since 1  T  is  equivalent to  [tex]V m^{-2} \cdot s[/tex]

         [tex]B_o = 2.0 4 *10^{-6} \ T[/tex]

Answer:

C. R^2

Explanation:

A cyclotron is a particle accelerator which employs the use of electric and magnetic fields for its functioning. It consists of two D shaped region called dees and the magnetic field present in the dee is responsible for making sure the charges follow the half-circle and then to a gap in between the dees.

R is denoted as the radius of the final orbit then the final particle energy is proportional to the radius of the two dees. This however translates to the energy being proportional to R^2.

Answer:

The charge on the bead is  [tex]q = 6.084 *10^{-7}\ C[/tex]

Explanation:

From the question we are told that

     The  mass of the bead is  [tex]m = 3.6 \ g = 0.0036 \ kg[/tex]

      The  magnitude of the electric field is  [tex]E = 200,000 \ N/C[/tex]

       The  acceleration of the bead is  [tex]a = 24 m/s^2[/tex]

Generally,  the  electric force on the bead is  mathematically represented as  

         [tex]F_ e = q E[/tex]

Where q is the charge on the bead

   Now the gravitational force opposing the upward movement of the bead is  mathematically represented as

       [tex]F_g = mg[/tex]

Generally the net force on the bead is  mathematically represented as

       [tex]F = F_e - F_g = m* a[/tex]

=>    [tex]qE - mg = ma[/tex]

Now substituting values

       [tex]q * 200000 - 0.0036 *9.8 = 0.0036 * 24[/tex]

     [tex]q = 6.084 *10^{-7}\ C[/tex]