**Electric Field**

Part A = 0.466 m

**Solution Below:**

_{1}= −31μC and Q

_{2}= 49μC, are separated by a distance of 12 cm. The electric field at the point P is zero.

Part A

_{1}is P?

The net electric field is given by the formula:

E(net) = (k * Q1 / r^{2}) + (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 / r^{2}) + (k * Q2 / (r + 0.12)^{2})

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

-(k * Q1 / r^{2}) = (k * Q2 / (r + 0.12)^{2})

k cancels from both sides:

-Q1 / r^{2} = Q2 / (r + 0.12)^{2}

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

-Q1 * (r + 0.12)^{2} = Q2 * r^{2}

Divide by Q2:

(-Q1 / Q2) * (r + 0.12)^{2} = r^{2}

This gives us:

-((-31μC) / (49μC)) * (r + 0.12)^{2} = r^{2}

0.63265 * (r + 0.12)^{2} = r^{2}

Distribute out the (r + 0.12)^{2}:

0.63265 * (r^{2} + 0.24r + 0.0144) = r^{2}

This gives:

0.63265r^{2} + 0.152r + 0.0091 = r^{2}

Get everything onto one side:

0 = 0.36765r^{2} – 0.152r – 0.0091

Solve the quadratic equation:

r = 0.466 m