Question

Two point charges lie on the x axis. A charge of + 2.20 pC is at the origin, and a charge of − 4.80 pC is at x=−11.0cm. What third charge should be placed at x=+20cm so that the total electric field at x=+10.0cm is zero?

Answer 1

As we know that electric field at any point caused due to a point charge given by, kQ/r^2

Where, k is dielectric constant of medium, Q is the point charge and r is the distance between the point charge and point where electric field is to be measured.

Let us assume a charge of q is placed at x=+20cm to make electric field at
x=+10cm zero.

Total electric field at x=+10cm can be shown by equation,

(k X 2.20 X 10^-12)/(10-0)^2 + (k X -4.80 X 10^-12)/(10-(-11))^2 + (k X
q)/(10-20)^2 = 0

=> 2.20 X 10^-14 - 1.09 X 10^-14 + q X 10^-2 = 0

=> q = 1.09 X 10^-12 - 2.20 X 10^-12

=> q = -1.1 X 10^-12 Coulombs or -1.1 pC

Answer 2

Answer: The charge must be -1.11pC

Explanation: The field done by a charge is equal to:

E(x) = k*q/(x - x')^2

Where k is the constant, q is the charge and x' is the position of the charge, then, the total field at the point x = 10cm must be also zero, we have:

E(10cm) = k*(2.20pC/(10cm - 0cm)^2 - 4.80pC/(10cm + 11cm)^2 + Q/(10cm - 20cm)^2) = 0

And we must find the value of Q.

(2.20pC/(10cm)^2 - 4.80pC/(21cm)^2 + Q/(-10cm)^2) = 0

Q = (-2.20pC/(10cm)^2 + 4.80pC/(21cm)^2)*(10cm)^2 = -1.11pC