Question

A charge of 2 micro coulomb is placed at x=0 and charge of -32 micro coulomb at x=60cm

Answer 1

Let q1 = 5 × 10–8 C and q2 = –3 × 10–8 C

Use the formula , Potential V = q/4πεor

Suppose Potential is zero at a distance x cm from q1 hence from charge q2 it is (16-x) cm

⇒ V = o

⇒ [q1/4πεox × 10–2] + [q2/4πεo(16-x)× 10–2] = 0

On solving we get , x = 0.1 m = 10 cm

Q.2. A regular hexagon of side 10 cm has a charge 5 μC at each of its
vertices. Calculate the potential at the centre of the hexagon.

Sol. charge q = 5 × 10–6 C

The distance of each charges from the centre of hexagon is r = 10 cm = 0.1 m

Hence Net potential at the centre is V = 6 ×(q/4πεor)

On putting the values we get , V = 2.7 × 10–6 volt

Q.3. Two charges 2 μC and –2 μC are placed at points A and B 6 cm
apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this
surface ?

Sol. (a) The plane normal to AB and passing through its mid point has zero potential everywhere .

(b) Normal to the plane in direction of AB

Answer 2

Answer:

Explanation:

Because of unlike charges the -Q charge will lie outside the other two points .