a ball with a mass of 0.150 g and positive charge of q=41.2 C is suspended on a string of negligible mass in a uniform electric field. We observe that the ball hangs at an angle of =22.0o from the vertical. What is the magnitude of the electric field?
force down on ball = m g
m = .15*10^-3 = 1.5*10^-4 kg
g = 9.81
I assume E is horizontal ?
Force sideways on ball = q E
tan 22 = qE / mg
To find the magnitude of the electric field, we can use the formula for the electric force on a charged object in an electric field. The electric force is given by the equation F = qE, where F is the electric force, q is the charge, and E is the electric field strength.
In this case, the electric force is balanced by the weight of the ball, causing it to hang at an angle from the vertical. The weight of the ball is given by the equation W = mg, where m is the mass of the ball and g is the acceleration due to gravity.
Since the ball is hanging at an angle from the vertical, we need to decompose the weight vector into its horizontal and vertical components. The vertical component of the weight, W_y, is given by W_y = W * sin(theta), where theta is the angle of the string with respect to the vertical. The horizontal component of the weight, W_x, is given by W_x = W * cos(theta).
Since the electric force is balanced by the vertical component of the weight, we have F = W_y. We can substitute the equations for F, W, and W_y into the equation F = qE to get qE = mg * sin(theta).
Now, we can solve for the electric field strength E by rearranging the equation as E = (mg * sin(theta)) / q.
Plugging in the given values:
m = 0.150 g = 0.150 * 10^-3 kg (convert grams to kilograms)
q = 41.2 C
theta = 22.0 degrees (convert degrees to radians by multiplying by pi/180)
Substituting these values into the equation, we get:
E = (0.150 * 10^-3 kg * 9.8 m/s^2 * sin(22.0 * pi/180)) / 41.2 C
Now, calculate the value using a calculator.