Question

To model this process, assume two charged spherical conductors are connected by a long conducting wire and a 1.20-mC charge is placed on the combination. One sphere, representing the body of the airplane, has a radius of 6.00 cm; the other, representing the tip of the needle, has a radius of 2.00 cm. (a) What is the electric potential of each sphere? (b) What is the electric

Answer 1

Part a: The electric potential of each sphere is 1.35x10⁸V

Part b: The electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively

Step-by-step explanation:

As the complete question is not given, the similar question is attached herewith. The values are used as indicated in the given question

Let r_1 = 6 cm=0.06 m

r2 = 2 cm = 0.02 m

Q = 1.2 mC

Let q1 and q2 are the charges on each sphere.

q1 + q2 = 1.2 mC -------(1)

In the equilibrium, V1 = V2

k*q1/r1 = k*q2/r2

q1/0.06 = q2/0.02

q1/q2 = 0.06/0.02

q1/q2 = 3 ---------(2)

On solving equation 1 and 2

we get

q1 = 0.9 mC

q2 = 0.3 mC

So

V1 = k*q1/r1 = (9*10^9*0.9*10^-3)/0.06 = 1.35*10^8 Volts

V2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02 = 1.35*10^8 Volts

So the electric potential of each sphere is 1.35x10⁸V

Part b

Now the electric potential is given as

E1 = k*q1/r1^2 = 9*10^9*0.9*10^-3/0.06^2 = 2.25*10^9 N/C

E2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02^2 = 6.75*10^9 N/C

So the electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively