Question

The particle P shown in figure (31-E11) has a mass of 10 mg and a charge of −0⋅01 µC. Each plate has a surface area 100 cm2 on one side. What potential difference V should be applied to the combination to hold the particle P in equilibrium?
Figure

Answer 1

Answer:

The particle is balanced when the electrical force on it is balanced by its weight.

Thus,

mg = qE mg = q×V'd ...(i)

Here,

d = Separation between the plates of the capacitor

V' = Potential difference across the capacitor containing the particle

We know that the capacitance of a capacitor is given by

C = ∈0A/d

⇒d=∈0A/C

Thus, eq. (i) becomes

mg = q×V'×C/∈0A

⇒V' = mg∈0A/q×C

⇒V' = 10−6×9.8×(8.85×10−12)×(100×10−4)/(0.01×10−6)×(0.04×10−6)

⇒V' = 21.68 mV

Since the values of both the capacitors are the same,

V = 2V' = 2 × 21.86 ≈ 43 mV

Answer 2

The potential difference of approximately 43 mV should be applied to hold the particle in equilibrium to the combination.

Explanation:

Step 1:

When the electric force is on it, the particle is balanced by the weight. So,  

$\mathrm{Mg}=\mathrm{gE} \mathrm{mg}=\mathrm{g} \times \mathrm{V}^{\prime} \mathrm{d}$ ...(1)

Where,  

d = Capacitor plates separation

V’ = Potential variance across the capacitor

Step 2:

It is to be understood that the capacitor’s capacitance is shown as  

$\Rightarrow d=\in 0 \mathrm{AC}$

Step 3:

So, the equation 1 is $\mathrm{Mg}=\mathrm{q} \times \mathrm{V}^{\prime} \times \mathrm{C} \in \mathrm{OA}$  

$\Rightarrow \mathrm{V}^{\prime}=10-6 \times 9.8 \times(8.85 \times 10-12) \times(100 \times 10-4)(0.01 \times 10-6) \times(0.04 \times 10-6)$

$\Rightarrow V^{\prime}=21.68 \mathrm{mV}$

Both the capacitors have the same value. $V=2 V^{\prime}=2 \times 21.86 \approx 43 \mathrm{mV}$