Question

Show that electric potential at a point P, at a distance 'r' from a fixed point charge Q, is given by:
    `v=(1/(4pi∈_0))Q/r`.

Answer 1

 

Let potential at point P is required due to a fixed point charge Q at A. Distance `bar(AP)`= r. Let us place a point test charge q₀ at P. Let us consider a point B very near to P at a distance dr, such that \[ \overrightarrow{F_P} = \overrightarrow{F_B}\].

Force, \[\overrightarrow{F}\] between Q and q₀ = `1/(4pi∈_0) -(Qxxq_0)/r^2`

where ∈₀ = permittivity for free space 

Now, work done in carrying the charge q₀ from P to B is 

                 `dW = -1/(4pi∈_0) (Qxxq_0)/r^2`.dr

The negative sign indicates negative work due to repulsion between Q and q_0.

Similarly,

Work was done in carrying to  infinity (∞),

                      W = `int dW=- int_r^∞ 1/(4pi∈_0). (Qxxq_0)/r^2`.dr

⇒                 `W =-1/(4pi∈_0) .Qxxq_0` `" _r^∞r^-2`dr

⇒                 `W = +1/(4pi∈_0) .Qxxq_0[1/r]_r^∞`

⇒                  `W = +1/(4pi∈_0) .Qxxq_0[1/∞ ∼1/r]`

⇒                  `W = -1/(4pi∈_0) .(Qxxq_0)/r`

When, the test charge, q₀ is taken from ∞.

The work done is positive and magnitude remains the same.

                  ∴  `W  =1/(4pi∈_0) .(Qxxq_0)/r`

Since potential at P = `W/(q_0)`

∴ V. potential at P = `1/(4pi∈_0) .(Qxxq_0)/(r.q_0)`

                         `V = Q/(4pi∈_0.r) J C^-1 or V`

                          `V = (1/(4pi∈_0)).Q/r V`