DivideByZero Posted April 28, 2009 Share Posted April 28, 2009 This is a hw problem so please explain your process instead of telling me the whole answer, thanks! There is a conductive ball hanging from a string in a electric plate. As you can see its slightly tilted toward the negative end. mass of ball: 0.04g potential difference: 480V separation of plates: 0.06m angle of string hanging: 20 degrees the question asks: What is the charge of the ball? I drew a free body diagram of three forces. one: force of tension two: gravity/weight three: force of charge pointing towards the right I think I'm suppose to add up the three forces to equal zero. F1 + F2 + F3 = 0. I think F3 should use the formula ©(q1)(q2)/(r^2). Am I right? If so then what is r??? I'm so lost in this problem please help guys! Link to comment Share on other sites More sharing options...
Klaynos Posted April 28, 2009 Share Posted April 28, 2009 If I were you I'd write down the force due to an electric field to start with... You should probably be familiar with the lorentz force http://en.wikipedia.org/wiki/Lorentz_force Or at least a very simplified version of it for a charge in a uniform electric field... Link to comment Share on other sites More sharing options...
DivideByZero Posted April 28, 2009 Author Share Posted April 28, 2009 If I were you I'd write down the force due to an electric field to start with... You should probably be familiar with the lorentz force http://en.wikipedia.org/wiki/Lorentz_force Or at least a very simplified version of it for a charge in a uniform electric field... electric field: E = F/q E * q = F (480)(q) = F holy s*** thanks for that hint! Merged post follows: Consecutive posts mergedOK so here is what I've done so far: F_g = m*g = (0.00004kg)*(9.8) = 3.92e-4 F_e = E*q = (480)*(q) F_t = m*g*cos(20) = 3.684e-4 F_g + F_e + F_t = 0 ... q = 1.58e-6 is this right????? Link to comment Share on other sites More sharing options...
NeonBlack Posted April 29, 2009 Share Posted April 29, 2009 electric field: E = F/q E * q = F (480)(q) = F No. Potential difference (voltage) is not the same as electric field. You will need to use the potential difference and the geometry of the plates to calculate the electric field. Link to comment Share on other sites More sharing options...
DivideByZero Posted April 29, 2009 Author Share Posted April 29, 2009 No. Potential difference (voltage) is not the same as electric field. You will need to use the potential difference and the geometry of the plates to calculate the electric field. oh so i have to use E=V/d and add the vectors... Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now