Question

Part A: A charge +Q is located at the origin and a second charge, +9Q, is located at x= 15.8 cm . Where should a third charge q be placed so that the net force on q is zero? Find q 's position on x -axis.
Part B: A charge +Q is located at the origin and a negative charge, -7Q, is located at a distance x= 19.6 cm . Where should a third charge q be placed so that the net force on q is zero? Find q 's position on x-axis.

Answer 1

Answer:

Part a)

x = 3.95 cm

Part b)

x = - 11.9 cm

Explanation:

Part a)

Since both charges are of same sign

so the position at which net force is zero between two charges is given as

$\frac{kq_1}{r_1^2} = \frac{kq_2}{(15.8 - r)^2}$

here we know that

$q_1 = Q$

$q_2 = 9Q$

$\frac{Q}{r^2} = \frac{9Q}{(15.8 - r)^2}$

square root both sides

$(15.8 - r) = 3r$

$r = 3.95 cm$

Part b)

Since both charges are of opposite sign

so the position at which net force is zero will lie on the other side of smaller charges is given as

$\frac{kq_1}{r_1^2} = \frac{kq_2}{(19.6 + r)^2}$

here we know that

$q_1 = Q$

$q_2 = -7Q$

$\frac{Q}{r^2} = \frac{7Q}{(19.6 + r)^2}$

square root both sides

$(19.6 + r) = 2.64r$

$r = 11.9 cm$

so on x axis it will be at x = - 11.9 cm