A small sphere of charge q = +68 ƒÝC and mass m = 5.8 g is attached to a light
string and placed in a uniform electric field E that makes an angle ƒá = 37„a with the horizontal. The opposite
end of the string is attached to a wall and the sphere is in static equilibrium when the string is horizontal as
in Figure P15.22.
(a) Construct a free body diagram for the sphere.
(b) Find the magnitude of the electric field.
(c) Find the tension in the string.
(a) To construct a free body diagram for the sphere, we need to analyze the forces acting on it.
1. The force of gravity, acting vertically downwards, can be represented by the weight of the sphere, mg.
2. The tension in the string, T, exerts an upward force on the sphere.
3. The electric field, E, exerts a horizontal force on the sphere in the direction opposite to its direction of inclination.
(b) To find the magnitude of the electric field, we need to consider the equilibrium condition. In static equilibrium, the net force on the sphere is zero.
1. The vertical component of the tension, Tsin(ć), balances the force of gravity: Tsin(ć) = mg.
2. The horizontal component of the tension, Tcos(ć), balances the force due to the electric field: Tcos(ć) = Eq.
Since we have values for q, m, ć, and g, we can substitute these values into the equations and solve for E.
(c) To find the tension in the string, we can use the equation Tsin(ć) = mg, which we derived in the previous step. We know the values of m, g, and ć, so we can substitute these values and calculate T.
Now let's substitute the given values and solve the equations.
(a) Free body diagram:
The free body diagram would show a small sphere with downward weight (mg), the tension force (T) acting vertically from the wall, and the electric field force (Eq) acting horizontally in the opposite direction of the angle of inclination.
(b) Magnitude of the electric field:
From the equation Tcos(ć) = Eq, we can rearrange it to find E:
E = Tcos(ć) / q.
Substituting the given values:
E = Tcos(37°) / 68 × 10^-9 C.
(c) Tension in the string:
From the equation Tsin(ć) = mg, we can solve for T:
T = mg / sin(ć).
Substituting the given values:
T = (5.8 × 10^-3 kg) × (9.8 m/s²) / sin(37°).
Now you can calculate the magnitude of the electric field (b) and the tension in the string (c) using the formulas and the given values.