Question

Two point charges of magnitude +7.00 μC and -9.00 μC are placed along the x-axis at x = 0 cm and x = 40.0 cm, respectively. Where must a third charge, q, be placed along the x-axis so that it does not experience any net force because of the other two charges?

Answer 1

Answer:

third charge must be placed at x = -298.11 cm

Explanation:

When two opposite charges are placed in a straight line then the electric field is always zero at a point near to the charge of lesser magnitude

so here let say the point where electric field is zero lie on the left side of 7.00 micro C charge at a distance "x" from it

so we will have net electric field due to both charges will be zero at that point

so electric field of charge 1 must be equal to the electric field of charge 2

$\frac{kq_1}{x^2} = \frac{kq_2}{(L + x)^2}$

so we have

$\frac{7\mu C}{x^2} = \frac{9 \mu C}{(40 + x)^2}$

now square root both sides

$\frac{2.645}{x} = \frac{3}{40 + x}$

$105.83 + 2.645 x = 3x$

$105.83 = 0.355 x$

$x = 298.11 cm$

so third charge must be placed at x = -298.11 cm