2. A point charge of +1.0 µC is moved in the direction of an electric field, and it has a change in electric potential difference energy of 10.0 J. What was the change in electric potential difference? 1 µC = 10-6 C
(Points : 1)
-1.0 × 10^8 V
-1.0 × 10^7 V
+1.0 × 10^7 V
+1.0 × 10^8 V
Question 3.3. A voltmeter reads 200.0 V between parallel plates that are 2.0 cm apart. What is the strength of the electric field between the two plates?
(Points : 1)
1.0 × 10^1 N/C
1.0 × 10^2 N/C
1.0 × 10^3 N/C
1.0 × 10^4 N/C
Volts + charge = work done
If + charge moves in direction of E field, potential goes down
.02 meters
E = V/d
To solve these questions, we need to understand the relationship between electric potential difference, electric field, and charge.
1. The change in electric potential difference energy is given by the equation:
ΔPE = qΔV
where ΔPE is the change in electric potential difference energy, q is the charge, and ΔV is the change in electric potential difference.
In this case, a point charge of +1.0 µC is moved in the direction of an electric field, and it has a change in electric potential difference energy of 10.0 J. We need to find the change in electric potential difference, ΔV.
Using the given equation:
ΔPE = qΔV
10.0 J = (1.0 × 10^(-6) C)ΔV
Solving for ΔV:
ΔV = 10.0 J / (1.0 × 10^(-6) C)
= 10^7 V
Therefore, the change in electric potential difference is +1.0 × 10^7 V.
2. The strength of the electric field between two parallel plates is given by the equation:
E = V / d
where E is the electric field, V is the potential difference between the plates, and d is the distance between the plates.
In this case, the voltmeter reads 200.0 V between the parallel plates that are 2.0 cm apart. We need to find the strength of the electric field, E.
Using the given equation:
E = V / d
E = 200.0 V / (2.0 × 10^(-2) m)
Converting cm to m:
E = 200.0 V / (0.02 m)
E = 10^4 N/C
Therefore, the strength of the electric field between the two plates is 1.0 × 10^4 N/C.