Question

Two charges –4 μC and 8 μC are 27 cm apart. The distance from the –4 μC charge on the line joining the charges where electric potential would be zero is?

Answer 1

Question

Two charges –4 μC and 8 μC are 27 cm apart. At what point the line joining the charges where electric potential would be zero?

Solution-

Giveb that, qA = -4 μC = -4 × 10^(-6)C, qB = 8μC = 8 × 10^(-6) C and distance between qA & qB is 27 cm (0.27 m = y)

Assume that there is a point "M" between qA and qB and distance between qA and M is "x" m. So, the distance between qB and M is (0.27 - x) m.

Potential at point M is the sum of potentials caused by charge qA and charge qB respectively.

V = qA + qB = 0

qA = - qB

qA/(4πεo × r) = -qB/(4πεo × r)

qA/x = -qB/(y - x)

Substitute the known values,

(4 × 10^-6)/x = -(-8 × 10^-6)/(0.27 - x)

(4 × 10^-6)/(x × 8 × 10^-6) = 1/(0.27 - x)

1/2x = 1/(0.27 - x)

Cross-multiply them,

2x = 0.27 - x

3x = 0.27

x = 0.09

Therefore, point M is at a distance of 0.09 m from charge qB.

Answer 2

Hey Pretty Stranger!

Here, I'm solving this Question using formula :

$\bigstar\large \sf \: \dfrac{kq_1}{r_1} = \dfrac{kq_2}{r_2}$

$\longrightarrow \sf \: \dfrac{4}{x} = \dfrac{8}{27 - x}$

$\longrightarrow \sf \: 4(27 - x) = 8 \times x$

$\longrightarrow \sf \: 108- 4x= 8x$

$\longrightarrow \sf \: - 4x - 8x = 108$

$\longrightarrow \sf \: - 12x = - 108$

$\longrightarrow \sf \: x = \cancel\dfrac{ - 108}{ - 12}$

$\longrightarrow \sf \: x = 9\: cm\: i.e. \: 0.09\: m$

Refer to the attachment ^^