Question

A) An isolated spherical conductor has an excess charge of -9.8 μC placed on its surface. Inside the conductor is a cavity, within which is a point charge of 4.10 μC. How many excess electrons are on the exterior surface of the conductor?

B) A cubical box has a charge q = -2.1 μC placed at its center. Calculate the electric flux through the right face of the box.

C) A charged ball with a mass of 10.0 g is suspended on a string in a horizontal electric field of 150.0 N/C directed to the right. The string makes an angle of 13.0 deg to the right of vertical. Find the charge carried by the ball.

D)Two charges, q₁ = -16.5 μC and q₂ = -1.5 μC , are located at (x,y) = (13.2, 22.2) cm and (10.6, 16.4) cm respectively. Find the electrostatic force between these two changes (take an attractive force to be negative)

Answer 1

(A) Charge inside the cavity, q = 4.10 uC

The charge on the spherical conductor Q = -9.8 uC. Now what happens is that the charge inside the cavity will attract an equal amount of opposite charge to the inner side of the spherical conductor (towards the cavity side) so that the electric field inside the conductor becomes zero. The attracted charge comes from the excess charge placed on the conductor.

Now, the excess charge on conductor becomes, Q1= |Q|-|q|= 5.7 uC

No. of excess electrons on the exterior surface = 5.7 uC/ charge of single electron = 5.7 uC/1.6x10^-19 C

= 3.56 x 10¹³ electrons approx.

(B) We use the Gauss's Law to find out the total electric flux of the cube. The charge enclosed Q is -2.1 uC

Flux through right face (just one face) = Total flux/6

(C) We will equate the forces acting on the ball in X and Y directions. The gravitational force acts in (-)Y-direction and the electrostatic force acts in (+)X-direction.

In Y-direction:

In X-direction:

(D) Electrostatic force is given as

The distance r between two charges is given as

Now the force is given as;