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?
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.
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
Kailey - I showed you how to do part c an hour ago.
a) To find the electric potential difference across a capacitor, you can use the formula:
V = Q/C
where V is the electric potential difference in volts (V), Q is the charge in coulombs (C), and C is the capacitance in farads (F).
In this case, you are given the charge as 2.1 × 10^-3 C and the capacitance as 5.2 µF. However, it is important to convert both values to the same units before using the formula.
To convert 5.2 µF to farads (F), you need to multiply by the conversion factor:
1 µF = 1 × 10^-6 F
So, 5.2 µF = 5.2 × 10^-6 F.
Now, you can substitute the values into the formula:
V = (2.1 × 10^-3 C) / (5.2 × 10^-6 F)
Simplify the expression:
V = 4.038 × 10^2 V
Therefore, the electric potential difference across the 5.2 µF capacitor with a charge of 2.1 × 10^-3 C is approximately 403.8 V.
b) (a) To find the charge on the oil drop, you can use the formula:
F = qE
where F is the weight of the drop in newtons (N), q is the charge on the drop in coulombs (C), and E is the electric field intensity in N/C.
In this case, you are given the weight as 8.5 × 10^-15 N and the electric field intensity as 5.3 × 10^2 N/C.
You can rearrange the formula to solve for q:
q = F / E
Substitute the given values:
q = (8.5 × 10^-15 N) / (5.3 × 10^2 N/C)
Simplify the expression:
q = 1.603 × 10^-17 C
Therefore, the charge on the oil drop is approximately 1.603 × 10^-17 C.
(b) To find the number of electrons the drop carries, you can use the fact that the elementary charge of an electron is 1.6 × 10^-19 C.
Divide the charge on the drop by the elementary charge:
Number of electrons = (1.603 × 10^-17 C) / (1.6 × 10^-19 C)
Simplify the expression:
Number of electrons = 10^2
Therefore, the oil drop carries approximately 100 electrons.
c) To find the electric potential difference between two plates, you can use the formula:
V = Ed
where V is the electric potential difference in volts (V), E is the electric field intensity in N/C, and d is the distance between the plates in meters (m).
In this case, you are given the electric field intensity as 7500 N/C and the distance between the plates as 0.05 m.
Substitute the given values into the formula:
V = (7500 N/C) × (0.05 m)
Simplify the expression:
V = 375 V
Therefore, the electric potential difference between the two plates is 375 V.