Question

1.) A 5.99?C and a -2.04?C charge are placed 16.2cm apart.

    

Where can a third charge be placed so that it experiences no net force? [Hint: Assume that the negative charge is 16.2cm to the right of the positive charge.

2.) A downward force of 3.0 N is exerted on a -8.4?C

charge.

a.) What is the magnitude of the electric field at this point?

b.)What is the direction of the electric field at this point?

Answer 1

if any charge +q is placed at any point between the two charges, it will be repelled by the positive charge and attracted by the negative charge resulting in force due to the two in the same direction and hence these two forces in the same direction cannot cancel.
If the charge +q is placed to the left of the positive charge, it will be repelled by the +ve charge and attracted by the negative charge, but the force of repulsion will be more than the attraction as +ve charge is > -ve charge and also the distance from the +ve charge is < than the distance from the -ve charge
=> +q charge must be placed to the right of the negative charge.
If x cm is the distance from the negative charge where there is no net force, then
k * 5.99q/(16.2 + x)^2 = k * 2.04q/x^2
=> (16.2 + x)/x = ?[(5.99)/(2.04)]
=> (16.2 + x)/x = 1.713
=> x = 22.70 cm.

b.procedure is to divide the force by the charge,

obtaining -357142 N/C.
for the "magnitude" of the field,
so you would drop the minus sign.
Final answer: 3.57 x 10^5 N/C

direction is opposite to force direction= upwards