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  • A 10 C charge is divided into two parts and placed at 1 cm distance so that the repulsive force between them

A 10 \muC charge is divided into two parts and placed at 1 cm distance so that the repulsive force between them
is maximum. The charges of the two parts are:
 

Option: 1

7 \mu C,3\mu C


Option: 2

8 \mu C,2\mu C


Option: 3

9 \mu C,1\mu C


Option: 4

5\mu C,5\mu C


Answers (1)

best_answer

Divide   q=10 \mu C into parts (x) and (q-x)
F=\frac{(K)(x)(q-x)}{r^2}For \ \mathrm{F}$ to be maximum $$ \begin{aligned} & \frac{d F}{d x}=0 \\ & x=\frac{q}{2}=\frac{10 \mu \mathrm{C}}{2}=5 \mu \mathrm{C} \\ & \mathrm{q}-\mathrm{x}=10 \mu \mathrm{C}-5 \mu \mathrm{C}=5 \mu \mathrm{C} \end{aligned}

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Ritika Harsh

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