At points P and Q, two identical charges each q are placed. When we move from P to Q (oa - the line joining them), electrostatic potential

Options

A. goes on decreasing
B. goes on increasing.
C. first goes on decreasing, then after becoming 1ninimum, it goes on increasing.
D. first increases, becomes maximum and then decreases.

Answer

A, B, C, D

Explanation

becoming minimum, it goes on increasing.

Above figure shows the situation as per the statement.
Consider a point A on the line segment `bar(PQ)` , at distance x from point P. Total electric potential at point A is,
`V= V1 +V_(2)`
`= (kq)/(x) +(kq)/(r-x)`
`:. V= kq((1)/(x)+(1)/(r-x))`
`:. (dV)/(dx)=kq[(-(1)/(x^(2)))-(1)/((r-x)^(2))(-1)]`
`:. (dV)/(dx)=kq [(1)/((r-x)^(2))-(1)/(x^(2))]`
Now taking `(dV)/(dx) =0` we get
`0 =kq[(1)/((r-x)^(2))-(1)/(x^(2))]`
`:. (1)/(r-x)^(2)=(1)/(x^(2))`
`:. x^(2)=(r-x)^(2)`
`:. x^(2)=r^(2)-2xr+x^(2)`
`:. 2xr = r^(2)`
`:. x =(r)/(2)`
`implies "At" x = (r)/(2) V ` = max . OR V = min.
Now `(d^(2)V)/(dx^(2))=(d)/(dx)((dV)/(dx))`
`:. (d^(2)V)/(dx^(2))= (d)/(dx) [kq{(1)/((r-x)^(2))-(1)/(x^(2))}]`
`=kq[-(2)/((r-x)^(3))(0-1)-(-(2)/(x^(3)))]`
`:. (d^(2)V)/(dx^(2))=2kq[(1)/((r-x)^(3))+(1)/(x^(3))]`
Now placing x = `(r)/(2)` in above equation
`((d^(2)V)/(dx^(2)))_("at" x=(r)/(2))=2kq[(8)/(r^(3))+(8)/(r^(3))]`
`= (32kq)/(r^(3))`
`gt0`
`implies "At" x = (r)/(2)` value of V is minimum.
Thus, as we move from P to C, potential goes on decreasing initially, then potential becomes minimum at mid point C and then potential goes on increasing.