Question

Two tiny spheres of mass 10 mg carry charges of equal magnitude, 69 nC, but opposite sign. They are tied to the same ceiling hook by light strings of length 508 mm. When a horizontal uniform electric field E that is directed to the left is turned on, the spheres hang at rest with the angle ? between the strings equal to 57° as seen in the figure below. What is the magnitude E of the field? (Your result must be in multiples of 103 N/C. That means if you get a result of a 9.22x103 just type 9.22 in the answer box. Include 2 digit after the decimal point and maximum of 5% of error is accepted in your answer. Take k=9x109 Nm2/C2 and g=9.81 m/s2).

Answer 1

From figure,

By force balance at equilibrium, on positive charge in vertical direction,

T*cosA = mg

given, = 57.0 deg

A = /2 = 28.5 deg (by symmetry)

m = mass of spheres = 10 mg = 10*10^-6 kg

g = 9.81 m/s^2

So, T = tension in rope = (10*10^-6)*9.81/cos(28.5 deg)

T = 11.16*10^-5 N

now by force balance at equilibrium, on positive charge in horizontal direction,

q*E = T*sinA + Fc

here, Fc = electrostatic force between spheres= k*q^2/r^2

here, r = separation distance = 2*L*sin(28.5 deg)

r = 2*0.508*sin(28.5 deg) = 0.485 m

given , q = 69.0 nC = 69.0*10^-9 C

So, Fc = [(9*10^9)*(69.0*10^-9)^2]/(0.485^2)

Fc = 1.82*10^-4 N

So, E = [(11.16*10^-5)*sin(28.5 deg) +1.82*10^-4]/(69.0*10^-9) = 3409.43 N/C

E = 3.41*10^3 N/C

Let me know if you've any query.