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Physics Question on electrostatic potential and capacitance

Two point-charges ±10μC\pm 10\mu C are placed 5.00 mm5.00\,mm apart, forming an electric dipole. Compute electric field at a point on the axis of the dipole 15 cm15\,cm away from the centre on a line passing through the centre dipole.

A

1.66×105NC−11.66\times {{10}^{5}}N{{C}^{-1}}

B

3.66×105NC−13.66\times {{10}^{5}}N{{C}^{-1}}

C

2.66×10−5NC−12.66\times {{10}^{-5}}N{{C}^{-1}}

D

2.66×105NC−12.66\times {{10}^{5}}N{{C}^{-1}}

Answer

2.66×10−5NC−12.66\times {{10}^{-5}}N{{C}^{-1}}

Explanation

Solution

The magnitude of the electric dipole moment is p=q×2l=(10×10−6)×(5.00×10−3)p=q\times 2l=(10\times 10^{-6})\times (5.00\times 10^{-3}) =50×10−9 cm=50\times 10^{-9}\,cm The field is required at point far-away from the centre of the dipole. The electric field at an axial point (end-on position) distant r(=15 cm)r(=15\,cm) from the centre of the dipole is E=14π ε02pr3=(9.0×109)2×(50×10−9)(15×10−2)3E=\frac{1}{4\pi \,\varepsilon }_{0}\frac{2p}{r^{3}}=(9.0\times 10^{9})\frac{2\times (50\times 10^{-9})}{(15\times 10^{-2})^{3}} 2.66×105NC−12.66 \times 10^5NC ^{-1}