Two point charges \( 5\times10^{-11}\,\mathrm{C} \) and \( -5\times10^{-11}\,\mathrm{C} \) are placed along x-axis at x = 3 cm and x = -3 cm respectively. What will be the electric field intensity at a point (0, 4) cm?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2.16\times10^{2}\,\mathrm{N/C} \)
The field at a point due to several charges is the vector sum of the fields of each charge (superposition principle); each field has magnitude \( kq/r^{2} \).
The point P(0, 4) cm is at the same distance from both charges:
\( \displaystyle r = \sqrt{3^{2} + 4^{2}}\,\mathrm{cm} = 5\,\mathrm{cm} = 0.05\,\mathrm{m} \)
\( \displaystyle E = \frac{kq}{r^{2}} = \frac{9\times10^{9} \times 5\times10^{-11}}{(0.05)^{2}}\,\mathrm{N/C} = \frac{0.45}{0.0025}\,\mathrm{N/C} = 180\,\mathrm{N/C} \)
The field of the positive charge at (3, 0) points away from it, towards upper left; the field of the negative charge at (-3, 0) points towards it, towards lower left. Their y-components cancel and their x-components add (both along \( -x \)). Each line from a charge to P makes an angle with the x-axis whose cosine is \( 3/5 \):
\( \displaystyle E_{\mathrm{net}} = 2E\cos\theta = 2 \times 180 \times \frac{3}{5}\,\mathrm{N/C} = 216\,\mathrm{N/C} \)
\( E_{\mathrm{net}} = 2.16\times10^{2}\,\mathrm{N/C}\ \text{(along } -x\text{)} \)
The point lies on the equatorial line of the dipole, so the net field is parallel to the axis, from the positive towards the negative charge.