Two point charges +8q and - 2q are located at x = 0 and x = L respectively. location of a point on x

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1. Electric Charges and Fields

medium

Two point charges $+8q$ and $ - 2q$ are located at $x = 0$ and $x = L$ respectively. The location of a point on the $x$-axis at which the net electric field due to these two point charges is zero is

$8 L$

$4 L$

$2 L$

$\frac{L}{4}$

(AIEEE-2005)

Solution

(c)The net field will be zero at a point outside the charges and near the charge which is smaller in magnitude.
Suppose $E.F.$ is zero at $P$ as shown.
Hence at $P$; $k.\frac{{8q}}{{{{(L + l)}^2}}} = \frac{{k.(2q)}}{{{l^2}}}$ $==>$ $l = L$.
So distance of $P$ from origin is $L + L = 2L$.

Standard 12

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Two point charges $+8q$ and $ - 2q$ are located at $x = 0$ and $x = L$ respectively. The location of a point on the $x$-axis at which the net electric field due to these two point charges is zero is

Solution

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