a. Q = C V
Therefore V = Q/C
Q is the charge in coulombs
C is the capacitance in farads, which in your case is 5.2*10^-6
V will be in volts
b. m g = Q E
E = 530 n/C
m g = 8.5810^-15 N
Q = n e where e is the electron charge, 1.602*10^-19 C
Solve for n.
The number of electrons, n, that you get should be a small integer, or nearly so.
a.What electric potential difference exists across a 5.2 µF capacitor that has a charge of 2.1 10-3 C?
how do i use uF and C into a formula to find answer??
b.An oil drop is negatively charged and weighs 8.5 10-15 N. The drop is suspended in an electric field intensity of 5.3 102 N/C.
(a) What is the charge on the drop?
(b) How many electrons does it carry?
c.The electric field intensity between two large, charged, parallel metal plates is 7500 N/C. The plates are 0.05 m apart. What is the electric potential difference between them?
3 answers
C = q/V
that is definition of capacitor, charge in coulombs over voltage in volts
5.2*10^-6 farads = 2.1 *10^-3 coulombs / v
so
V = 2.1 * 10^-3 / 5.2 *10^-6
= .404 * 10^3
= 404 volts
that is definition of capacitor, charge in coulombs over voltage in volts
5.2*10^-6 farads = 2.1 *10^-3 coulombs / v
so
V = 2.1 * 10^-3 / 5.2 *10^-6
= .404 * 10^3
= 404 volts
Kailey - I showed you how to do part c an hour ago.
1 answer