Answer
a) $0.294$ m
b) $8.82\times 10^{-7}C$
Work Step by Step
We can find the required distance and charge as follows:
(a)
$V_1=\frac{Kq}{r_1}$
$\implies V_1r_1=Kq$......eq(1)
Similarly $V_2=\frac{Kq}{r_2}$
$\implies V_2r_2=Kq$......eq(2)
Comparing eq(1) and eq(2), we obtain:
$V_1r_1=V_2r_2$
$\implies V_1r=V_2(r+1m)$
We plug in the known values to obtain:
$(2.7\times 10^4V)r=6140V(r+1m)$
This simplifies to:
$r=0.294m$
(b)
From eq(1), we have
$q=\frac{V_1r_1}{K}$
We plug in the known values to obtain:
$q=\frac{(2.7\times 10^4V)(0.294m)}{8.99\times 10^9N.m^2/C^2}$
$q=8.82\times 10^{-7}C$
