Poteintial difference to move mass

[indigojoker]

There's a small sphere of mass m that hangs by a thread. the sphere is between two parallel plates L apart. The sphere has a charge Q. What is the potential difference that will make the plates to assume an angel of 20 degrees with the vertical?

F=QE=mg \sin \theta
E=\frac{mg \sin 20}{Q}
V= - \int _0 ^L \frac{mg \sin 20}{Q}dx
V= - \frac{mgL \sin 20}{Q}d

i feel like this question should involve more work. was my thought process correct?

 

[Avodyne]

Yes, it's correct. (But the d in your last line is presumably a typo.) Also, since the question asks for "potential difference", the minus sign is not really relevant.

 

[indigojoker]

yes it's a typo

could you explain why the potential difference is not relevant?

 

[Andrew Mason]

There's a small sphere of mass m that hangs by a thread. the sphere is between two parallel plates L apart. The sphere has a charge Q. What is the potential difference that will make the plates to assume an angel of 20 degrees with the vertical?

F=QE=mg \sin \theta
E=\frac{mg \sin 20}{Q}
V= - \int _0 ^L \frac{mg \sin 20}{Q}dx
V= - \frac{mgL \sin 20}{Q}d

i feel like this question should involve more work. was my thought process correct?

How do you get the first statement? Do a free body diagram of the forces on the suspended mass. There is the tension in the string, gravity and the electric force. From a free body vector diagram you should be able to get the expression for E and then V (=E/L) .

AM

 

[Andrew Mason]

If you do the free body diagram you will see that the gravitational and electric forces have to be balanced by the tension in the string.

\vec{T} = q\vec{E} + m\vec{g}

This means that:

T\sin{20} = qE
T\cos{20} = mg

Dividing:

qE/mg = \tan{20}

EL = V, so

V = mgL \tan{20}/q

AM