Question

2 41 92 observation point

(a) Using the symmetry of the arrangement, determine the direction of the force on q in the figure above, given that qa = qb = −7.00 µC and qc = qd = +7.00 µC.

(b) Calculate the magnitude of the force in newtons on the charge q, given that the square is 10.0 cm on a side and q = 1.00 µC.

Answer 1

(a) From an above figure, we have

The x-component of force from q_a on charge 'q' which acts as "RIGHT".

The x-component of force from q_b on charge 'q' which acts as "LEFT".

The x-component of force from q_c on charge 'q' which acts "RIGHT".

The x-component of force from q_d on charge 'q' which acts "LEFT".

The horizontal forces cancel out as they are equal in magnitude and opposite in direction. The vertical component of each force is in the downward direction. Thus, the net force is in downward direction.

Using the symmetry of an arrangement, the direction of force on charge 'q' in above figure will be given as -

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(b) Distance from any corner to charge 'q' = (1/2) (diagonal of the square)

d = (1/2) _$\sqrt{}$[(0.1 m)² + (0.1 m)²]

d = (1/2) (0.1414 m)

d = 0.0707 m

Total force exerted by each charge on 'q' which will be given as -

F = k_e |q_a| |q| / d²

F = [(9 x 10⁹ Nm²/C²) (7 x 10^-6 C) (1 x 10^-6 C)] / (0.0707 m)²

F = 12.6 N

Therefore, magnitude of the force on charge 'q' which will be given by -

| F | = 4 F cos 45⁰

| F | = [4 (12.6 N) (0.7071)]

| F | = 35.6 N