Question: Q. 1. Five charges, $q$ each are placed at the corners of a regular pentagon of side $a$.
(i) What will be the electric field at $O$ if the charge from one of the corners (say $A$ ) is removed ?
(ii) What will be the electric field at $O$ if the charge $q$ at $A$ is replaced by $-q$ ?
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Solution:
Ans. (i) $\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r^{2}}$ along $O A$
(ii) $\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 q}{r^{2}}$ along $O A$
[CBSE Marking Scheme, 2017]
Commonly Made Error
- Some candidates find resultant force $F$ as a scalar sum of all forces, but resultant force is equal/to vector sum of all forces.
Detailed Answer :
(i) If a charge $q$ is removed frompoint $A$, a negative charge is developed at $A$ where electric field will be $E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r^{2}}$ which is along $\mathrm{OA}$.
(ii) If a charge $q$ is replaced by charge $-q$ at point $A$, there generates a net electric field at point $O$ as a result of $-2 q$ charge, so
$$ \begin{aligned} & E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{-2 q}{r^{2}} \ & E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 q}{r^{2}} \text { along } O A \end{aligned} $$
