Question

A positive charge Q1 = 34 nC is located at the origin. A negative charge Q2 = -5.5 nC is located on the positive x-axis p = 15 cm from the origin. Calculate the location of x axis in cm where electric field is.

Answer 1

Answer:

10.7 cm and 25.1 cm

Explanation:

Given:

Q_1 = 34 * 10^(-9) C

Q_1 = -5.5 * 10^(-9) C

Distance between charges = 0.15 m

Distance from origin where E is zero = x

k = 8.99*10^9 Nm^2 / C^2

It is to be noted that E due to positive charge at origin is directed along + x-axis while for negative charge it is directed towards the origin. General, the electric field strength E at a distance R due to charge q is given by :

E = k*(Q) / R^2

The Net Electric field strength of the two charges and is equated to zero as follows:

E_net = k*(Q_1) / x^2 + k*(Q_2) / (0.15 - x)^2 = 0

(Q_1) / x^2 + (Q_2) / (0.15 - x)^2 = 0

Solve the above equation for x:

Q_1 * (0.15 - x)^2 + Q_2 * x^2 = 0

Q_1 * (x^2 - 0.3x + 0.0225) - Q_2 * x^2 = 0

(Q_1 + Q_2) x^2 - 0.3*Q_1*x + 0.0225*Q_1 = 0

2.85x^2 - 1.02x + 0.0765 = 0

x = 0.107 m

x = 0.251 m