Question

(II) Determine the direction and magnitude of the electric field at the point P in Fig. 16–56. The charges are separated by a distance 2a, and point P is a distance x from the midpoint between the two charges. Express your answer in terms of Q, x, a, and k.

Short answer

The magnitude of the net electric field at point P is \(\frac{{ - 4kQxa}}{{{{\left( {{x^2} - {a^2}} \right)}^2}}}\), and it is directed toward the left.

Step-by-step solution

Understanding of electric field

The electric field at a particular point due to a point charge is in the plane containing the point and the charge. The electric field obeys the principle of superposition as the field is linear related to the charge.

Evaluation of the direction and magnitude of the electric field at the point P

The schematic diagram for the problem can be drawn as follows:

Consider the right direction to be positive. Then the field due to \( + Q\) will be positive, and the field due to \( - Q\) will be negative.

The electric field at point P due to charge \( - Q\) is calculated as follows:

\({E_1} = - \frac{{kQ}}{{{{\left( {x - a} \right)}^2}}}\)

The electric field at point P due to charge \( + Q\) is calculated as follows:

\({E_2} = \frac{{kQ}}{{{{\left( {x + a} \right)}^2}}}\)

The magnitude of the net electric field at point P can be calculated as follows:

\(\begin{aligned}{c}E = {E_1} + {E_2}\\E = - \frac{{kQ}}{{{{\left( {x - a} \right)}^2}}} + \frac{{kQ}}{{{{\left( {x + a} \right)}^2}}}\\E = kQ\left( {\frac{1}{{{{\left( {x + a} \right)}^2}}} - \frac{1}{{{{\left( {x - a} \right)}^2}}}} \right)\\E = kQ\left( {\frac{{\left( {{x^2} + {a^2} - 2ax} \right) - \left( {{x^2} + {a^2} + 2ax} \right)}}{{{{\left( {{x^2} - {a^2}} \right)}^2}}}} \right)\\E = \frac{{ - 4kQxa}}{{{{\left( {{x^2} - {a^2}} \right)}^2}}}\end{aligned}\)

Here, the negative sign shows the net field directed toward the left.

Thus, the magnitude of the net electric field at point P is \(\frac{{ - 4kQxa}}{{{{\left( {{x^2} - {a^2}} \right)}^2}}}\), and it is directed toward the left.