Name: Dipole in an Electric Field
Uploaded: 2019-07-19
Duration: 11 min 59 s
Description: Dipoles exists everywhere. They apply to both electricity and magnetism in physics. Here in this video, let us explore how an electric dipole interacts with...

Question

A charge $A$ of $8.0 μC$ is situated at a distance of $1.00$ metre from a charge $B$ of $- 2.0 μC$ , as shown. Calculate the potential at the mid-point $C$ between $A$ and $B$ . Also calculate the potential at $D$ which is at a distance of $0.80 m$ from $A$ and $0.20 m$ from $B$ . How much work will be done in taking $0.02 μC$ of charge from $D$ to $C$ ?

Accepted Answer

Given that

$8.0 μC$

Potential at $c$ is

$V_{C} = V_{A} + V_{B} = \frac{1}{4 π \in_{0}} \frac{q_{1}}{AC} + \frac{1}{4 π \in_{0}} \frac{q_{2}}{BC} = \frac{9 \times 10^{9} \times 8 \times 10^{- 6}}{0.50} + \frac{9 \times 10^{9} \times (- 2 \times 10^{- 6})}{0.50} \frac{9 \times 10^{9} \times 10^{- 6}}{0.50} [8 - 2] = \frac{9 \times 10^{9} \times 10^{- 6} \times 6 \times 100}{50} V_{C} = 1.08 \times 10^{5} V olt$

Potential at D --

$V_{D} = V_{A} + V_{B} = \frac{1}{04 π \in_{0}} \frac{q_{1}}{AD} + \frac{1}{4 π \in_{0}} \frac{q_{2}}{BD} = \frac{9 \times 10^{9} \times 8 \times 10^{- 6}}{0.80} - \frac{9 \times 10^{9} \times 2 \times 10^{- 6}}{0.20} = 9 \times 10^{9} \times 10^{- 6} [\frac{8}{0.80} - \frac{2}{0.20}] = 9 \times 10^{3} [\frac{800}{80} - \frac{200}{20}] = 9 \times 10^{3} \times 0 = 0$

Work done from $D$ to $C$ $\to$

$W = q (V_{C} - V_{D}) = 0.02 \times 10^{- 6} \times 1.08 \times 10^{5} = 2.16 \times 10^{- 3} \Rightarrow W_{DC} = 2.16 \times 10^{- 3} J$

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