Question

Two point charges +5 C and -2 C are kept at a distance of 1 m in free space. The distance between the two
zero potential points on the line joining the charges is:

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Answer 1

Answer:2/7

Explanation:equating potentials of both the charge

Answer 2

Concept: The charges in our agreement have been cancelled, so we have a potential of zero. The potential is zero exactly midway between two identical and oppositely charged point charges, for example. It takes no energy to move a particle between any two points of the same potential (zero or not).

Given: Two point charges +5C and -2C

            distance between them = 1m

To find: Distance between the two zero potential points on the line joining the charges

Solution:

Let the potential be 0 at P and Q.

For $x_{1}$

$\frac{k*5}{x_{1} } = \frac{2k}{(1-x_1)}$

$(k*5) (1-x_1) = 2kx_1\\5k - 5kx_1 = 2kx_1\\5k = 2kx_1 + 5kx_1\\5k = 7kx_1$

$x_{1} = \frac{5}{7}$

In the same way,

$\frac{k*5}{1+x_2} = \frac{k*2}{x_2}\\(5kx_2) = 2k(1+x_2)\\5kx_2 = 2k + 2kx_2\\3kx_2 = 2k\\x_2 = \frac{2}{3}$

Distance between P and Q

= $(1 - \frac{5}{7} ) + \frac{2}{3}\\\frac{7-5}{7} + \frac{2}{3} \\\frac{2}{7} + \frac{2}{3}$

LCM of the denominators is 21

$\frac{6+14}{21}\\\frac{20}{21}$

Hence, the distance between the two zero potential points on the line joining the charges is $\frac{20}{21} m$.

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