Question

Given that a positive charge Q1=+3c and a negative charge Q2=-2c,are separated by a distance of 8cm.Take permittivity constant k=9×10E9Nm2/C2.
i.find the electric field at a point p,1cm from Q2(m)
ii.what is the electric potential at a point p,5cm from Q1.

Answer 1

i) The electric field at that point is the superposition of the individual charge fields. The field from $Q_1$ is given as follows:

$$E_1 = \dfrac{kQ_1}{r_1^2}$$

where $r_1 = 8cm - 1cm = 7cm = 0.07m$ is the distance from the first charge to the point.

Similarly:

$$E_2 = \dfrac{kQ_2}{r_2^2}$$

where $r_2 = 1cm = 0.01m$ is the distance from $Q_2$ to the point.

The total field is:

$$E = E_1 + E_2 = \dfrac{kQ_1}{r_1^2} + \dfrac{kQ_2}{r_2^2} = k\left( \dfrac{Q_1}{r_1^2} + \dfrac{Q_2}{r_2^2} \right)\\ E = 9\times 10^9\cdot \left( \dfrac{3}{0.07^2} - \dfrac{2}{0.01^2} \right) \approx -1.7\times 10^{14}N/C$$

ii) Similarly:

$$E = k\left( \dfrac{Q_1}{r_1^2} + \dfrac{Q_2}{r_2^2} \right)$$

where now $r_1 = 5cm = 0.05m$ is the distance from $Q_1$ to the point, and $r_2 = 8cm - 5cm = 0.03m$ is the distance from $Q_2$ to the point. Substituting the numbers, obtain:

$$E = 9\times 10^9\cdot \left( \dfrac{3}{0.05^2} - \dfrac{2}{0.03^2} \right) = -9.2\times 10^{12}N/C$$

Answer. i) $-1.7\times 10^{14}N/C$, ii) $-9.2\times 10^{12}N/C$.