Question

Two point charges q1=2.1*10^(-8) and q2=-4q1 are fixed in two places 50 cm apart.

The point charge q1 is at the right side of q2. Find the point at which the electric field is zero, along the straight line passing through the two charges. The point is?

a)50cm to the right of q1

b)50cm to the left of q2

c)100cm to the right of q2

d)100 cm to the left of q2

Answer 1

Electric field is given by:

E = kQ/R^2

Since we know that direction of electric field due to +ve charge will be away from charge and due to -ve charge will be towards the charge. In given case since q1 charge is positive, So electric field will be away from q1 and since q2 is negative so, electric field will be towards the charge.

Now Suppose q2 is at origin then, q1 will be at x = +50 cm. So from this we can see that electric field will be zero towards the right of origin. Suppose at distance d from origin, Electric field is zero then

Enet = E1 - E2 = 0

E1 = E2

kq2/r2^2 = kq1/r1^2

q1 = 2.1*10^-8 C

q2 = -4*q1 = -8.4*10^-8 C

r2 = d

r1 = d - 50 cm = d - 0.5

Using above values:

8.4*10^-8/d^2 = 2.1*10^-8/(d - 0.5)^2

d/(d - 0.5)= sqrt (8.4/2.1) = 2

d = 2*(d - 0.5)

d = 2*d - 1

d = 1 m = 100 cm

d = 100 cm to the right of q2 = (100 - 50) cm to the right of q1

d = 50 cm to the right of q1

Correct option is A.

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