First slide

Electric field

Question

Two point charges of $20 μC$ and $80 μC$ are 10 cm apart. Where will the electric field strength be zero on the line joining the charges from $20 μC$ charge

Easy

A

0.1 m
B

0.04 m
C

0.033 m
D

0.33 m

Solution

Let neutral point be obtained at a distance x from $20 μC$ charge. Hence at neutral point
$\frac{20}{(x)^{2}} = \frac{80}{(10 - x)^{2}} \Rightarrow x = + 0 .033 m$

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