Example

Question:

Two point charges \( q_1 \) and \( q_2 \), of magnitude \( +10^{-8}\,\text{C} \) and \( -10^{-8}\,\text{C} \), respectively, are placed \( 0.1\,\text{m} \) apart. Calculate the electric fields at points A, B and C shown in Fig. 1.11.

Solution:

The electric field vector \( \mathbf{E}_{1A} \) at A due to the positive charge \( q_1 \), points towards the right and has a magnitude \[ E_{1A} = \frac{(9 \times 10^9\,\text{Nm}^2\text{C}^{-2}) \times (10^{-8}\,\text{C})}{(0.05\,\text{m})^2} = 3.6 \times 10^4\,\text{N\,C}^{-1} \] The electric field vector \( \mathbf{E}_{2A} \) at A due to the negative charge \( q_2 \) points towards the right and has the same magnitude. Hence the magnitude of the total electric field \( \mathbf{E}_A \) at A is \[ E_A = E_{1A} + E_{2A} = 7.2 \times 10^4\,\text{N\,C}^{-1} \] \( E_A \) is directed toward the right.

The electric field vector \( \mathbf{E}_{1B} \) at B due to the positive charge \( q_1 \) points towards the left and has a magnitude \[ E_{1B} = \frac{(9 \times 10^9\,\text{Nm}^2\text{C}^{-2}) \times (10^{-8}\,\text{C})}{(0.05\,\text{m})^2} = 3.6 \times 10^4\,\text{N\,C}^{-1} \] The electric field vector \( \mathbf{E}_{2B} \) at B due to the negative charge \( q_2 \) points towards the right and has a magnitude \[ E_{2B} = \frac{(9 \times 10^9\,\text{Nm}^2\text{C}^{-2}) \times (10^{-8}\,\text{C})}{(0.15\,\text{m})^2} = 4 \times 10^3\,\text{N\,C}^{-1} \] The magnitude of the total electric field at B is \[ E_B = E_{1B} - E_{2B} = 3.2 \times 10^4\,\text{N\,C}^{-1} \] \( E_B \) is directed towards the left.

The magnitude of each electric field vector at point C, due to charge \( q_1 \) and \( q_2 \) is \[ E_{1C} = E_{2C} = \frac{(9 \times 10^9\,\text{Nm}^2\text{C}^{-2}) \times (10^{-8}\,\text{C})}{(0.10\,\text{m})^2} = 9 \times 10^3\,\text{N\,C}^{-1} \] The resultant is \[ E_C = E_{1C} \cos\left(\frac{\pi}{3}\right) + E_{2C} \cos\left(\frac{\pi}{3}\right) = 9 \times 10^3\,\text{N\,C}^{-1} \] \( E_C \) points towards the right.