Two fixed charges `-2Q` and `+Q` are located at points `(-3a,0)` and `(+3a,0)` respectively. Then which of the following statement is correct?

Answer

a. centre at (5 a, 0) and radius `4 a`
b. `(##BMS_V03_C03_E01_056_A01##)`
c. `sqrt((1)/(4 pi epsilon_0)((Q q)/(2 m a)))`

Explanation

a. Let `P` be a point in the `xy` plane with coordinates `(x, y)` at which the potential due to charges `-2 Q and + Q` placed `A` and `B`, respectively, be zer (Fig. S3.24).
(##BMS_V03_C03_E01_056_S01##).
`V_P = (K (- 2 Q))/sqrt((3 a + x )^2 + y^2) + (K(+ Q))/sqrt((3 a - x)^2 + y^2)= 0`
or `2 sqrt((3 a - x)^2 + y^2) = sqrt((3 a + x)^2 + y^2)`
or `4[(3 a - x)^2 + y^2 ] = [(3 a + x)^2 + y^2]`
or `(x - 5a)^2 + (y - 0)^2 = (4 a)^(2)`
This is the equation of a circle with center at `(5 a, 0)` and radius `4 a`. Thus, `c (5 a, 0)` is the center of the circle, where potential is zero.
b. Let us find potential at various points on the x - axis.
For `x gt 3a`,
`V = (k Q)/(x - 3a) - (k 2Q)/(x + 3 a) = (k Q( 9a - x))/(x ^2 - 9a^2)`
We see that at `x = 9a, V = 0`
For `-3a lt x lt 3a,`
`V = (kQ)/(3 a - x) - (k 2 Q)/(x + 3a) = (k Q 3(x - 1))/(9 a^2 - x ^2)`
We see that at `x = a, V = 0`.
For `x lt - 3a`,
`V = (kQ)/(3 a - x) + (k (-2 Q))/(-3 a - x) = (-k Q(9a -x))/(x^2 - 9 a^2)`
We see that `V ne 0` for `x lt - 3 a` and V is always negative for `x lt -3 a`.
c. Applying energy conservation, we get
`(KE + PE)_("center") = (KE + PE)_("circumference")`
or `0 + K [(Q q)/(2 a) - (2 Q q)/(8 a)] = (1)/(2) mv^2 + k[(Qq)/(6 a) -(2 Q q)/(12 a)]`
or `(1)/(2) mv^2 = (K Q q)/(4 a)` or `v = sqrt((K Q q)/(2 m a)) = sqrt((1)/(4 pi epsilon_0)((Q q)/(2 m a)))`
.