Two charges of magnitude +q and -3q are place | vedclass

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Question

(A) Let the distance from the charge $+q$ where the potential is zero be $x$ (in meters).The total potential $V$ at this point is the sum of the poten

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) Let the distance from the charge $+q$ where the potential is zero be $x$ (in meters).The total potential $V$ at this point is the sum of the potentials due to both charges:$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{x} + \frac{-3q}{1-x} ight) = 0$Since $\frac{1}{4 \pi \varepsilon_0} 
eq 0$,we have:$\frac{q}{x} - \frac{3q}{1-x} = 0$$\frac{1}{x} = \frac{3}{1-x}$$1 - x = 3x$$1 = 4x$$x = \frac{1}{4} \, m = 0.25 \, m = 25 \, cm$.Thus,the distance is $25 \, cm$.

Two charges of magnitude $+q$ and $-3q$ are placed $100 \, cm$ apart. The distance from $+q$ between the charges where the electrostatic potential is zero is ....... $cm$.

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