ElectrostaticsHard

Question

Two shorts dipoles $(A,B),A$ having charges $\pm 2\mu C$ and length 1 cm and $B$ having charges $\pm 4\mu C$ and length 1 cm are placed with their centres 80 cm apart as shown in the figure. The electric field at a point $P$, equi-distant from the centres of both dipoles is $\_\_\_\_$ N/C.

Options

A.$\frac{9}{16}\sqrt{2} \times 10^{5}$
B.$4.5\sqrt{2} \times 10^{4}$
C.$9\sqrt{2} \times 10^{4}$
D.$\frac{9}{16}\sqrt{2} \times 10^{4}$

Solution

$${{\overrightarrow{E}}_{2} = - \frac{{KP}_{2}}{r^{3}};{\overrightarrow{E}}_{1} = - \frac{2{KP}_{1}}{r^{3}} }{P_{1} = 2 \times 10^{- 6} \times 10^{- 2} = 2 \times 10^{- 8} }{P_{2} = 4 \times 10^{- 6} \times 10^{- 2} = 4 \times 10^{- 8} }{{\overrightarrow{E}}_{net}\frac{2 \times 9 \times 10^{9} \times 2 \times 10^{- 8}}{(0.4)^{3}}\widehat{i} - \frac{9 \times 10^{9} \times 4 \times 10^{- 8}}{(0.4)^{3}}\widehat{j} }{{\overrightarrow{E}}_{\text{net~}} = \frac{9 \times 10^{9} \times 4 \times 10^{- 8}}{(0.4)^{3}}\lbrack\widehat{i} - \widehat{j}\rbrack }{\left| {\overrightarrow{E}}_{\text{net~}} \right| = \frac{9 \times 10^{4}}{16}(\sqrt{2})}$$

Create a free account to view solution

View Solution Free
Topic: Electrostatics·Practice all Electrostatics questions

More Electrostatics Questions