6. If 4.0 × 1018 J of work done on a point charge changes its electric potential difference by 25.0 V, then what is the magnitude of the charge?
(Points : 1)
1.6 × 101^9 C
4.0 × 101^9 C
1.6 × 101^8 C
4.0 × 101^8 C
7. Two parallel plates have an electric field strength between them of 1.0 × 103 N/C and an electric potential difference of 50.0 V. How far apart are the plates?
(Points : 1)
1.0 cm
2.0 cm
5.0 cm
10.0 cm
Voltage * charge = work done by E field on charge
Same thing really
Volts = E * d
Use METERS volts, Coulombs
To find the magnitude of the charge in question 6, we can use the formula for electric potential difference:
Electric Potential Difference = Work Done / Charge
We are given that the work done is 4.0 × 10^18 J and the electric potential difference is 25.0 V. Plugging these values into the formula:
25.0 V = 4.0 × 10^18 J / Charge
To find the charge, we rearrange the formula:
Charge = 4.0 × 10^18 J / 25.0 V
Simplifying the expression:
Charge = (4.0 × 10^18 J) / (25.0 V)
To perform this calculation, we divide 4.0 × 10^18 by 25.0:
Charge = 1.6 × 10^17 C
Therefore, the correct answer for question 6 is:
1.6 × 10^17 C
For question 7, we can use the formula for electric field strength:
Electric Field Strength = Electric Potential Difference / Distance
We are given that the electric field strength is 1.0 × 10^3 N/C and the electric potential difference is 50.0 V. Plugging these values into the formula:
1.0 × 10^3 N/C = 50.0 V / Distance
To find the distance, we rearrange the formula:
Distance = 50.0 V / (1.0 × 10^3 N/C)
Simplifying the expression:
Distance = 50.0 V / 1.0 × 10^3 N/C
To perform this calculation, we divide 50.0 by 1.0 × 10^3:
Distance = 0.05 m
Since the distance is given in meters, we convert it to centimeters:
0.05 m = 0.05 × 100 cm = 5.0 cm
Therefore, the correct answer for question 7 is:
5.0 cm