Gujarati
Hindi
1. Electric Charges and Fields
hard

$5$ charges each of magnitude $10^{-5} \,C$ and mass $1 \,kg$ are placed (fixed) symmetrically about a movable central charge of magnitude $5 \times 10^{-5} \,C$ and mass $0.5 \,kg$ as shown in the figure given below. The charge at $P_1$ is removed. The acceleration of the central charge is [Given, $\left.O P_1=O P_2=O P_3=O P_4=O P_5=1 m , \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9\right]$

A

$9 \,ms ^{-2}$ upwards

B

$9 \,ms ^{-2}$ downwards

C

$4.5 \,ms ^{-2}$ upwards

D

$4.5 \,ms ^{-2}$ downwards

(KVPY-2009)

Solution

(c)

Forces on charge at point $O$ initially balances each other as it is given that acceleration occurs when charge at point $P_1$ is removed. This means resultant of force due to charges at points $P_2, P_3, P_4$ and $P_5$ is equal and opposite to force due to at point $P_1$.

Hence, acceleration of charge at point $O$ is directed along $O P_1$.

Acceleration $=\frac{F}{m}$

$=\frac{(\frac{K q_1 q_2}{r^2})}{m}$

$=\frac{9 \times 10^9 \times 10^{-5} \times 5 \times 10^{-5}}{(1)^2 \times 1}$

$=4.5 \,ms ^{-2}$

Standard 12
Physics

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