Two point charges of $+ 16 μ C$ and $- 9 μ C$ are placed $8 c m$ apart in air. Determine the position of the point at which the resultant field is zero.

Open in App

Solution

$E_{1} = \frac{k q_{1}}{x^{2}} = \frac{k \times q}{x^{2}}$
$E_{2} = \frac{k q_{2}}{(8 + x)^{2}} = \frac{k \times 16}{(x + 8)^{2}}$
$E_{1} = E_{2}$
$\frac{9 k}{x^{2}} = \frac{16 k}{(x + 8)^{2}}$
$\frac{3}{x} = \frac{4}{x + 8}$
$4 x = 3 x + 24$
$x = 24$ .

Suggest Corrections

Similar questions

Two fixed point charges $+ 16 μ C$ and $- 9 μ C$ are placed $8 c m$ apart in air. Where should a third point charge placed so that, the third charge is in equilibrium

Q. two point like charges having magnitude

+ 16 μ C

and

- 9 μ C

are separated by a distance

10 c m

in air. The resultant electric field will be zero.

Q. TWO point charges

+ 2 n c

and

- 4 n c

are

1 m

apart in air. Find the positions along the line joining the two charges at which resultant potential is zero.

Two electric charges of $9 μ C$ and $- 3 μ C$ are placed $0.16 m$ apart in air. There will be a point P at which electric potential is $z e r o$ on the line joining the two charges and in between them. The distance of P from $9 μ C$ charge is:

Two point charges 2 μC and 8μC are placed 12 cm apart. The position of point from 2 μC charge, where the electric field intensity is zero is:

Join BYJU'S Learning Program

Two point charges of +16μC and −9μC are placed 8 cm apart in air. Determine the position of the point at which the resultant field is zero.

E1=kq1x2=k×qx2E2=kq2(8+x)2=k×16(x+8)2E1=E29kx2=16k(x+8)23x=4x+84x=3x+24x=24.

Two point charges of $+ 16 μ C$ and $- 9 μ C$ are placed $8 c m$ apart in air. Determine the position of the point at which the resultant field is zero.

$E_{1} = \frac{k q_{1}}{x^{2}} = \frac{k \times q}{x^{2}}$
$E_{2} = \frac{k q_{2}}{(8 + x)^{2}} = \frac{k \times 16}{(x + 8)^{2}}$
$E_{1} = E_{2}$
$\frac{9 k}{x^{2}} = \frac{16 k}{(x + 8)^{2}}$
$\frac{3}{x} = \frac{4}{x + 8}$
$4 x = 3 x + 24$
$x = 24$ .