<h2>
Complete Question:</h2>
Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by 45.0 mm, and the potential difference between them is 360V.
a) What is the magnitude of the electric field (assumed to be uniform) in the region between the plates?
b) What is the magnitude of the force this field exerts on a particle with charge +2.40nC?
c) use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower.
d) Compare the result of part (c) to the change of potential energy of the same charge, computed from the electrical potential. Please walk me step and detail.
Source: https://www.chegg.com/homework-help/questions-and-answers/two-large-parallel-metal-plates-carry-opposite-charges-equal-magnitude-separated-450-mm-po-q770630
<h2>
Answer:</h2>
(a) 8 x 10³ V/m
(b) 19.2 x 10⁻⁶ N
(c) 8.64 x 10⁻⁷ J
(d) - 8.64 x 10⁻⁷ J
<h2>
Explanation:</h2>
(a) The electric field (E) between two parallel plates is related to the potential difference (V) and the distance (d) between the plates as follows;
V = E / d
=> E = V / d --------------------------(i)
<em>From the question, the following are given;</em>
V = 360V
d = 45.0mm = 0.045m
<em>Substitute these values into the equation (i) as follows;</em>
=> E = 360 / 0.045
=> E = 8000 V/m
Therefore, the magnitude of the electric field between the plates is 8000V/m or 8 x 10³ V/m
(b) The magnitude of the force (F) exerted on a charge (Q) by an electric field (E) is given by;
F = Q x E -----------------------(ii)
<em>From the question, the following are given;</em>
Q = charge on the particle = 2.40nC = 2.40 x 10⁻⁹C
E = electric field calculated above = 8000V/m
<em>Substitute these values into equation (ii) as follows;</em>
F = 2.40 x 10⁻⁹ x 8000
F = 19.2 x 10⁻⁶N
F = 19.2μN
Therefore, the magnitude of the force this field exerts on that particle is 19.2 μC
(c) The field moves from higher potential (positive plate) to a lower potential (negative plate) and thus the work done (W) in moving is related to the force (F) exerted on the particle and the distance (d) between the plates as follows;
W = F x d -----------------(iii)
<em>Known;</em>
F = 19.2 x 10⁻⁶ N
d = 0.045m
Substitute these values into equation (iii) as follows;
W = 19.2 x 10⁻⁶ x 0.045
W = 8.64 x 10⁻⁷ J
Therefore, the work done by the field is 8.64 x 10⁻⁷ or 0.864μ J
(d) The particle moves from higher potential (say V₂) to a lower potential (V₁) and has a potential difference of 360V i.e V₂₁ = V₂ - V₁ = 360V
Now to calculate the change in potential energy (ΔU) ,which is the potential difference from lower to higher potential of the charge Q (i.e ΔV = -360V), the following relation is used;
ΔU = Q x ΔV
ΔU = 2.40 x 10⁻⁹C x -360
ΔU = - 8.64 x 10⁻⁷ J
Therefore, the change in potential energy (moving from lower to higher potential) is the negative of the work done by the field. i.e
ΔU = - 8.64 x 10⁻⁷ J = -W
