Answer:
The work done by the force is 6μJ
Explanation:
Suppose, A 30 nC charge is moved from a point where V₁ = 130 V to a point where V₂ = -70 V.
We need to calculate the work done by the force
Using formula of work done
Where, q = charge
=potential difference
Put the value into the formula
Hence, The work done by the force is 6μJ.
(e) halved
<h2>Explanation:</h2>The electrical enery (E) stored in a capacitor is related to its capacitance (C) and potential difference (V) as follows;
E = x C x
------------------------(i)
Also, the capacitance (C) of a capacitor consisting of parallel plates is related to the area (A) of the plates and distance (d) between the plates as follows;
C = A x ε₀ / d ------------------------(ii)
Where;
ε₀ is the permittivity of free space.
Substituting equation (ii) into equation (i) gives;
E = x A x ε₀ / d x
--------------------(iii)
From equation(iii)
When the potential difference (V) is constant, then the electrical energy (E) stored is <em>inversely </em>proportional to the distance between the plates. i.e
E = k / d ----------------(iv)
Where;
k = proportionality constant = x A x ε₀ x
(which is the product of all constants)
Therefore from equation (iv);
=> E₁ x d₁ = E₂ x d₂ ---------------------------(v)
Where;
E₁ and E₂ are the initial and final values of the electrical energy stored.
d₁ and d₂ are the initial and final values of the distance between the plates.
<em>So, when the distance is doubled, i.e.</em>
d₂ = 2 x d₁
<em>Substitute the value of d₂ into equation (v) to give;</em>
=> E₁ x d₁ = 2 x d₁ x E₂
<em>Divide through by d₁ to give;</em>
=> E₁ = 2 x E₂
<em>Make E₂ subject of the formula</em>
=> E₂ = x E₁
Therefore, the electrical energy stored in the capacitor will be halved.