A uniform electric field of magnitude 1.0×10^2 NC^−1 exists between two conducting plates, one of which is positively charged and the other of which is negatively charged. The plates are 10 mm apart.
(a) Calculate the magnitude of the potential difference between the plates.
(b) If a proton is moved from the positive plate to the negative plate, what is the magnitude of the change in its electrostatic potential energy?
(c) If this device operates as a capacitor, how is it able to store electrostatic potential energy?
a) dV/dr = E
=> V = E*r
=> V = 10^2 * 10^-2m
= 1V
b) Change in potential energy = U = qV
Charge on a proton = q = 1.6*10^-19
U = (1.6*10^-19)*V
= (1.6*10^-19)*1
= 1.6*10^-19J
(a) Oh, conducting plates and electric fields! Plate jokes incoming! The magnitude of the potential difference between the plates can be calculated using the formula V = Ed, where V is the potential difference, E is the electric field, and d is the distance between the plates. So, V = (1.0×10^2 NC^−1) × (10 mm) converted to meters = 10 volts. That's electrifying!
(b) When the proton is moved from the positive plate to the negative plate, its electrostatic potential energy changes. To calculate the magnitude of this change, we can use the formula ΔPE = qΔV, where ΔPE is the change in electrostatic potential energy, q is the charge of the proton, and ΔV is the potential difference. Since the proton has a charge of 1.6×10^-19 C and the potential difference is 10 volts, we have ΔPE = (1.6×10^-19 C) × (10 volts) = 1.6×10^-18 joules. That's a tiny energy change, but protons are small!
(c) Ah, the capacitor! This device is able to store electrostatic potential energy because the electric field between the plates does work on charges that are moved from one plate to the other. When charges are separated against the electric field, work is done, and this work gets stored as potential energy. So, it's like the plates are holding a secret potential energy stash, just waiting to be used! Capacitors are like the piggy banks of electricity!
(a) The potential difference (V) between the plates can be calculated using the formula:
V = Ed,
where E is the magnitude of the electric field and d is the distance between the plates.
Given:
E = 1.0×10^2 NC^−1,
d = 10 mm = 10 × 10^(-3) m.
Substituting these values into the formula:
V = (1.0×10^2 NC^−1) × (10 × 10^(-3) m)
= 1.0×10^2 V.
So, the magnitude of the potential difference between the plates is 1.0×10^2 V.
(b) The change in electrostatic potential energy (ΔPE) of a charged particle moving in an electric field can be calculated using the formula:
ΔPE = qV,
where q is the charge of the particle and V is the potential difference.
Given:
q = charge of a proton = 1.6 × 10^(-19) C,
V = potential difference = 1.0×10^2 V.
Substituting these values into the formula:
ΔPE = (1.6 × 10^(-19) C) × (1.0×10^2 V)
= 1.6 × 10^(-19) J.
So, the magnitude of the change in electrostatic potential energy of the proton is 1.6 × 10^(-19) J.
(c) The device described operates as a capacitor. A capacitor is a device that stores electrical energy in the form of electrostatic potential energy. In this case, the uniform electric field between the plates creates a potential difference. When charges are moved from one plate to the other, work is done against the electric field and potential energy is stored in the electric field between the plates. The ability to store electric potential energy is a result of the separation of charges and the presence of the electric field created by the charges on the plates.
(a) To calculate the magnitude of the potential difference between the plates, we can use the formula:
ΔV = Ed
Where ΔV is the potential difference, E is the magnitude of the electric field, and d is the distance between the plates.
Given:
E = 1.0×10^2 NC^−1
d = 10 mm = 10×10^(-3) m
Substituting the values into the formula, we have:
ΔV = (1.0×10^2 NC^−1) * (10×10^(-3) m)
= 1.0 V
Therefore, the magnitude of the potential difference between the plates is 1.0 V.
(b) The change in electrostatic potential energy can be calculated using the formula:
ΔPE = qΔV
Where ΔPE is the change in electrostatic potential energy, q is the charge of the particle, and ΔV is the potential difference.
Given:
q = charge of a proton = +1.6×10^(-19) C
ΔV = 1.0 V
Substituting the values into the formula, we have:
ΔPE = (+1.6×10^(-19) C) * (1.0 V)
= +1.6×10^(-19) J
Therefore, the magnitude of the change in electrostatic potential energy of the proton is +1.6×10^(-19) J (Joules).
(c) A capacitor is able to store electrostatic potential energy by virtue of its ability to store charge on its plates. In this device, the positively charged plate and the negatively charged plate create an electric field between them. When a charge, such as a proton, is moved from the positive plate to the negative plate, work is done against the electric field. This work converts into an increase in the electrostatic potential energy of the proton. The capacitor stores this energy, which can later be released when the charges are discharged or used in an electrical circuit. Essentially, the capacitor acts as a temporary reservoir for electrostatic potential energy.