Question

Four charged rods form the side of a square in the horizontal (xy) plane. Each rod has a length 25.2 cm and each carries a uniformly distributed positive charge Q. A small sphere, which can be considered to be a point charge of mass 3.29 ✕ 10-4 g and electric charge +2.42 ✕ 10-12 C is in equilibrium at a location z = 21.5 cm above the center of the square. Find the value of Q. HINT: Don't ignore gravity in this case.

Answer 1

length of rod = l = 25.2 cm = 25.2X 10^(-2)mrs { l/2 = 12.6 X 10^(-2) Mrs}

Charge on each rod = +Q

Mass of point charge = M = 3.29 X 10^(-4) gms = 3.29 X 10^(-7) Kg

Charge on point charge + +2.42 X 10^(-12) C

co-ordinates of point charge = [0 ,0 , 21.5X10^(-2)] mrs

Call the position of the sphere P. Call the midpoint of any rod A and the midpoint of the opposite rod B.

Draw diagram showing triangle APB.

The distance, R (=AP = BP), from the center of a rod to the sphere is given by Pythagoras therom:
R² = (L/2)² + z²
. . .= 0.126^2² + 0.215²
R = 0.2492m

The angle between AP and AB is θ where tanθ = z/(L/2)
θ = tan⁻¹(z/(L/2))
. .= tan⁻¹(0.215/0.126)
. .= 59.63°

Note the mass = 3.29x10⁻⁴ g = 3.29x10⁻⁷ kg
______________________________

At the position of the small sphere, each rod produces a field radially outwards of magnitude:
E = (kQ/R)/√[(L/2)² + R²]
I hope this formula is derived in your class room.
E = (9.0x10⁹Q/0.2492) /√[0.126² + 0.2492²]
. .= 1.29x10¹¹Q
The force from each rod on the sphere (charge q) is
F = qE
. .= 1.29 x10¹¹Q x 2.42x10⁻¹²
. .= 0.312Q

The upwards vertical component of each force is Fsinθ. The 4 upwards vertical components balance the weight (mg). (Horizontal components cancel due to symmetry.)

4Fsinθ = mg
4 x 0.312Q x sin(59.62) = 3.29x10⁻⁷ x 9.81
Q = 3.00X10^(-6) C