Mastering Physics Solutions: Electric Field

Electric Field

Part A = 0.466 m

Solution Below:

Two point charges, Q1 = −31μC and Q2 = 49μC, are separated by a distance of 12 cm. The electric field at the point P is zero.

Part A

How far from Q1 is P?

The net electric field is given by the formula:

E(net) = (k * Q1 / r2) + (k * Q2 / (r + x)2)

r is the distance from point P. The problem tells us that E(net) is 0, and x, the distance between the two charges, is 12cm (convert this to 0.12m):

0 = (k * Q1 / r2) + (k * Q2 / (r + 0.12)2)

Subtract one of the two charges’ expressions (we used Q1):

-(k * Q1 / r2) = (k * Q2 / (r + 0.12)2)

k cancels from both sides:

-Q1 / r2 = Q2 / (r + 0.12)2

To simplify, multiply each side by the other’s denominator:

-Q1 * (r + 0.12)2 = Q2 * r2

Divide by Q2:

(-Q1 / Q2) * (r + 0.12)2 = r2

This gives us:

-((-31μC) / (49μC)) * (r + 0.12)2 = r2
0.63265 * (r + 0.12)2 = r2

Distribute out the (r + 0.12)2:

0.63265 * (r2 + 0.24r + 0.0144) = r2

This gives:

0.63265r2 + 0.152r + 0.0091 = r2

Get everything onto one side:

0 = 0.36765r2 – 0.152r – 0.0091

Solve the quadratic equation:

r = 0.466 m

Tagged with: