Recent content by mli273

Nevermind, figured it out on my own.

[b]1. Two capacitors, one 7.2 \muF the other 16 \muF, are connected in series across a 15 V battery. Find the charge stored on 7.2 \muF capacitor. Find the charge stored on 16 \muF capacitor. [b]2. C-eq =5.0\muF and V=Q/C and \epsilon=Q(1/C-eq) [b]3. I know that both Q values...

3x10^6= V/.0023, so V = 6900 Volts. Thank you, I always have trouble on the simple ones.

1. What plate area is required if an air-filled, parallel-plate capacitor with a plate separation of 2.3 mm is to have a capacitance of 24 pF? Which I found correctly to be 6.2 x 10^-3 m^2 by using the formula C= k(8.85x10^-12)A/d What is the maximum voltage that can be applied to this...

[b]1. Two point charges of equal magnitude are 8.4 cm apart. At the midpoint of the line connecting them, their combined electric field has a magnitude of 45 N/C. Find the magnitude of the charges. [b]2. E=k(q)/r^2 [b]3. I've tried 45= 8.99x10^9(q)/(.084)^2 and got q to equal 35...

that's true, I didn't think of it like that. So would it be 1-.915 which is .085. Thank you!

[b]1. A person riding in a boat observes that the sunlight reflected by the water is polarized parallel to the surface of the water. The person is wearing polarized sunglasses with the polarization axis vertical. If the wearer leans at an angle of 17.0 degrees to the vertical, what fraction...

Oh..I get it. So it would be 2q -mg= ma, so the answer would be 9.81 m/s^2 & it's positive because the force is upward. Thank you so much!

Ok, well a = [2(-2.9x10^-6)(4.1x10^4) - (.012)(9.81)]/.012 = -29.6 m/s^2 But this is not the correct answer according to Mastering Physics.

Net force was F=mg so F=(.012)(9.81)= .11772 N

is the net force doubled as well?

[b]1. An object with a charge of -2.9 microC and a mass of .012 kg experiences an upward electric force, due to a uniform electric field, equal in magnitude to its weight. Part A: Find the magnitude of the electric field. Which I correctly calculated to be 4.1 x 10^4 N/C. Part B: If the...