Electric Field Problem and simple pendulum

[bodensee9]

Homework Statement

If you had a simple pendulum of length 1 m and mass 5e-3 kg placed in a uniform electric field E that is directed vertically upward. The bob has charge of -8e-6 C. the period is 1.2 s what are the magnitude and direction of E?

First, didn't they already tell us that E is directed vertically upward, so wouldn't the direction of E be vertically upward? Though wouldn't there be a force downward on the charge from E if that is the case?

Also, I thought that the period for the simple pendulum is sqrt(L/g), where L is the length measured from the pivot and g is gravity. So here sqrt(L/g) doesn't come out to be 1.2 s?

Would the force from the Field qE = the torque that causes it to oscillatte (but can we ignore gravity?) And since F = ma, so we know that qE/m = a. And we also know that the angular acceleration on a pendulum is Lmgsin(theta)/I, where I is the moment of inertia of the pendulum. So does this mean that the two are equal (after I multiply the angular acceleration by L)? But then what about theta?

Thanks.

 

[Dick]

Yes, if the field points upward the force points downward. So its effect is to make the downward force on the object larger than mg. Just use sqrt(L/g) and make 'g' larger.