Answer:
V= -3.6*10⁻¹¹ V
Explanation:
Since the charge is uniformly distributed, outside the sphere, the electric field is radial (due to symmetry), so applying Gauss' Law to a spherical surface at r= 30 cm, we can write the following expression:[tex]E* A = \frac{Q}{\epsilon_{0} } (1)[/tex]
At r= 0.3 m the spherical surface can be written as follows:[tex]A = 4*\pi *r^{2} = 4*\pi *(0.3m)^{2} (2)[/tex]
Replacing (2) in (1) and solving for E, we have:[tex]E = \frac{Q}{4*\pi *\epsilon_{0}*r^{2} } = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)^{2} y} (3)[/tex]
Since V is the work done on the charge by the field, per unit charge, in this case, V is simply:V = E. r (4)Replacing (3) in (4), we get:[tex]V =E*r = E*(0.3m) = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)} = -3.6e11 V (5)[/tex]
V = -3.6*10¹¹ Volts.The electrical potential module will be [tex]-3.6*10^-^1^1 V[/tex]
We can arrive at this answer as follows:
To answer this, we owe Gauss's law. This is because the charge is evenly distributed across the sphere. This will be done as follows:[tex]E*A=\frac{Q}{^E0} \\\\\\A=4*\pi*r^2[/tex]
Solving these equations will have:[tex]E=\frac{Q}{4*\pi*^E0*r^2} \\E= \frac{(9e9N*m2/c2)*(-12C)}{(0.3m)^2y}[/tex]
As we can see, the electric potential is carried out on the field charge. In this case, using the previous equations, we can calculate the value of V as follows:[tex]V=E*r\\V=E*0.3m= \frac{(9e9N*m^2/C2)*(-12C)}{0.3m} \\V= -3.6*10^-^1^1 V.[/tex]
More information about Gauss' law at the link:
https://brainly.com/question/14705081
He throws a second ball (B2) upward with the same initial velocity at the instant that the first ball is at the ceiling. c. How long after the second ball is thrown do the two balls pass each other? d. When the balls pass each other how far are they above the juggler’s hands? e. When they pass each other what are their velocities?
Answer:
hello your question has some missing parts
A juggler performs in a room whose ceiling is 3 m above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.
answer : c) 0.39 sec
d) 2.25 m
e) 1.92 m/sec
Explanation:
The initial velocity of the first ball = 7.67 m/sec ( calculated )
Time required for first ball to reach ceiling = 0.78 secs ( calculated )
Determine how long after the second ball is thrown do the two balls pass each other
Distance travelled by first ball downwards when it meets second ball can be expressed as : d = 1/2 gt^2 = 9.8t^2 / 2
hence d = 4.9t^2 ----- ( 1 )
Initial speed of second ball = first ball initial speed = 7.67 m/sec
3 - d = 7.67t - 4.9t ---- ( 2 )
equating equation 1 and 2
3 = 7.67t therefore t = 0.39 sec
Determine how far the balls are above the Juggler's hands ( when the balls pass each other )
form equation 1 ;
d = 4.9 t^2 = 4.9 *(0.39)^2 = 0.75 m
therefore the height the balls are above the Juggler's hands is
3 - d = 3 - 0.75 = 2.25 m
determine their velocities when the pass each other
velocity = displacement / time
velocity = d / t = 0.75 / 0.39 sec = 1.92 m/sec