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Question: Two charges \(\pm 20\mu C\)are placed 10 mm apart. The electric field at point P, on the axis of the...

Two charges ±20μC\pm 20\mu Care placed 10 mm apart. The electric field at point P, on the axis of the dipole 10 cm away from its centre O on the side of the positive charge is.

A

8.6×109NC−18.6 \times 10^{9}NC^{- 1}

B

4.1×106NC−14.1 \times 10^{6}NC^{- 1}

C

3.6×106NC−13.6 \times 10^{6}NC^{- 1}

D

4.6×105NC−14.6 \times 10^{5}NC^{- 1}

Answer

3.6×106NC−13.6 \times 10^{6}NC^{- 1}

Explanation

Solution

:

Here, q=±20μC=±20×10−6Cq = \pm 20\mu C = \pm 20 \times 10^{- 6}C

Or 2a=10mm=10×10−3m2a = 10mm = 10 \times 10^{- 3}m

r=OP=10cm=10×10−2mr = OP = 10cm = 10 \times 10^{- 2}m

∣P→∣=q×2a=20×10−6×10×10−3m\left| \overrightarrow{P} \right| = q \times 2a = 20 \times 10^{- 6} \times 10 \times 10^{- 3}m

=2×10−7m= 2 \times 10^{- 7}m

The electric field along BP, , E=2P→r4πε0(r2−q2)2E = \frac{2\overrightarrow{P}r}{4\pi\varepsilon_{0}(r^{2} - q^{2})^{2}}

As a < < r,

E→=2∣P→∣4πε0r3=2×2×10−7×9×109(10×10−2)3=3.6×106NC−1\overrightarrow{E} = \frac{2\left| \overrightarrow{P} \right|}{4\pi\varepsilon_{0}r^{3}} = \frac{2 \times 2 \times 10^{- 7} \times 9 \times 10^{9}}{(10 \times 10^{- 2})^{3}} = 3.6 \times 10^{6}NC^{- 1}