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The force between two point charges q(1)...

The force between two point charges `q_(1) and q_(2)` separted by a distance r is `F = (kq_(1)q_(2))/(r^(2))` where k is a constant. Find the potential energy of the system of charges.

Text Solution

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`F = - (dU)/(dr) rArr dU = - Fdr`
Integrating `U = - int_(0)^(r) Fdr = -kq_(1)q_(1) int_(0)^(r) r^(-2)` dr
`rArr U = (kq_(1)q_(2))/(r)`
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Knowledge Check

  • Coulomb's electrostatic force between two charges in SI system is given by F = (Xq_(1)q_(2))/r^(2) where the value of x is

    A
    4piepsilon_(o)
    B
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  • Two equal charges q are placed at a distance of 2a and a third charge -2q is placed at the midpoint. The potential energy of the system is ……………

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    C
    `-(7q^(2))/(8piin_(0)a)`
    D
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  • The distance between two charges q_(1)=+2 muC and q_(2)+8 muC is 15 cm. Calculate the distance from the charge q_(1) to the points on the line joining the two charges where the electric field is zero

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