Question

# Two charges $$Q_1 = 18 \mu C$$ and $$Q_2 = - 2 \mu C$$ are separated by a distance R, and $$Q_1$$ is on the left of $$Q_2$$. The distance of the point where the net electric field is zero is :

A
between Q1 and Q2
B
left of Q1 at R/2
C
right of Q2 at R
D
right of Q2 at R/2

Solution

## The correct option is D right of $$Q_2$$ at R/2A. between $$Q_1$$ and $$Q_2$$ , electric field due to both charges will be in same direction so net electric field can not be zero in that direction. Hence A is wrongB.left of Q1 electric field dut to $$Q_1$$ will always be higher than $$Q_2$$ because$$Q_1\geq Q_2 and r_1\leq r_2$$. Hence B is wrongC. $$\displaystyle \frac{kQ_2}{x^2} = \frac{kQ_1}{(x + R)^2}$$ or $$x \displaystyle = \frac{R}{2}$$ Therefore C is wrong and D is correctPhysicsNCERTStandard XII

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