Q. 44.4( 8 Votes )

# Two equal charges are placed at a separation of 1.0 m. What should be the magnitude of the charges so that the force between them equals the weight of a 50 kg person?

Answer :

__Given:__

Mass of the person

Weight of the person,

Let the magnitude of charge on both be q.

Here,

Distance of separation,

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**Formula used:**

By Coulomb’s law, the electric force is given by:

Where ϵ_{0} is the permittivity of free space

k is the electrostatic constant

q_{1} and q_{2} are the magnitude of charges

r is the distance of separation between the charges

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Rearranging, we get

Substituting the corresponding values, we get

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charges q and –3q are placed fixed on x-axis separated by distance ‘d’. Where should a third charge 2q be placed such that it will not experience any force?

Physics - ExemplarNaCl molecule is bound due to the electric force between the sodium and the chlorine ions when one electron of sodium is transferred to chlorine. Taking the separation between the ions to be 2.75 × 10^{–8} cm, find the force of attraction between them. State the assumptions (if any) that you have made.