University of Wisconsin Green Bay

A 1.8 g (0.0018 kg) pith ball is suspended from a thread and placed in a 4500 N/C electric field. The field is directed horizontally to the left. When placed in the electric field, the pith ball is deflected to the left so that the thread makes a 15o angle with the vertical. How much charge is on the pith ball?

  • In this problem, you are asked to relate motion (the pith ball is at rest, 150 from vertical) to force (it is deflected when in the presence of an electric field). Force and motion of a single object are always related through Newton’s Second Law, so this is a force or 2nd Law problem.

    Nowhere in the statement of the problem are the words “force” or “acceleration” used. This is a case where your experience helps a lot! Whenever you have an equilibrium situation (in this case, static equilibrium) you know that acceleration is zero and Newton’s Second Law is a useful way to approach the problem. In addition, you know that the presence of an electric field means there is an electric force on a charged object, and the deflection of the pith ball is caused by that force.

  • Step 1:

    Your FBD is not yet finished, because T has both x- and y- components. Scroll down when you are ready to continue.


    Step 2:

    In the final FBD drawn here, all forces are divided into components. The contribution each force makes in the x-direction is shown explicitly, as is the contribution each force makes in the y-direction. The FBD is now a visual representation of ∑F=ma in each direction.

  • The key equation for any problem that relates forces and motion is Newton’s Second Law. Regardless of what quantity you are asked to find, begin with the Second Law. If additional information is needed, it will become apparent as you proceed.

  • The problem asks how much charge is on the pith ball, so no further mathematical solution is needed. Because the deflection is in the same direction as the electric field (both to the left,) you know that the charge on the ball is positive. (FE = qE)

    It does not matter whether you begin with the x- or the y-equation. If you solve for x first, you will need to pause in your solution to obtain a value for T from the y-equation.

  • In this problem, the pith ball doesn’t fall—the vertical component of tension supports it against gravity. Knowing the geometry of the problem allows you to solve for tension in the thread directly.


    When the ball is placed in the electric field, it will swing to the left until the electric force is balanced by a horizontal component of tension.

    Tension is a passive force—its value adjusts to the other forces present and to the motion of the object. Because tension must always have a vertical component to balance the weight of the ball, the tension in the rope will get larger as the ball swings to the left. (T2 = Tx2 + Ty2 = Tx2 + mg2, and Tx grows until qE is balanced.)