NEET · Physics Electrostatics Capacitors

Two quick reasoners: geometry of connection, and dielectric drop

Neither needs heavy algebra — both turn on a single relationship. Spot it and each is a five-second answer.

13 Series vs parallel of 7 identical capacitorsConnection geometry

Question

Seven identical capacitors connected in series give an equivalent capacitance $C$. What is the equivalent capacitance when the same seven are connected in parallel?

Series of 7 → C = c⁄7

Parallel of 7 → 7c

(a) C/49
(b) C/7
(c) 7C
(d) 49C

Given

Number7 identical caps
Each valuec (unknown)
Series equivalentC

Asked

The parallel equivalent $C_P$, expressed in terms of the given series value $C$.

Concept

For $n$ identical capacitors of value $c$: series gives $\dfrac{c}{n}$ and parallel gives $nc$. The ratio of parallel to series is therefore $n^2$.

So whatever the series value is, the parallel value is $n^2$ times larger. Here $n=7\Rightarrow n^2=49$.

Method & Steps

Find each cap from the series fact. $C_S=\dfrac{c}{7}=C \Rightarrow c=7C$.
Combine in parallel. $C_P = 7c = 7(7C)$.
Simplify. $$C_P = 49C$$

Easy Trick

Parallel ÷ series = n². For $n$ identical capacitors the parallel value is always $n^2$ times the series value — no need to find $c$ at all. With $n=7$: $C_P = 49\,C$.

(Resistors are the mirror image: there it's the series total that's $n^2$ times the parallel total.)

Solution

$c=7C$ from the series condition, so parallel gives $7c$:

$$C_P = 7 \times 7C = 49C$$

d 49 C parallel equivalent

14 Dielectric slab and the potential dropDielectric constant

Question

A parallel plate capacitor has capacitance $16\,\mu\text{F}$. When a glass slab is inserted between the plates, the potential difference falls to $\tfrac{1}{8}$ of its original value. Find the dielectric constant $K$ of the glass.

Charge stays fixed; the slab raises C by K, so V drops by K.

(a) 4
(b) 8
(c) 16
(d) 32

Given

Original C16 μF
New PDV′ = V⁄8
ChargeQ constant (isolated)

Asked

The dielectric constant $K$ of the glass slab.

Concept

The capacitor is disconnected, so the charge Q is fixed. A dielectric multiplies capacitance: $C' = K\,C$. With $Q$ constant and $V=\dfrac{Q}{C}$,

$$V' = \frac{Q}{C'} = \frac{Q}{K\,C} = \frac{V}{K}$$

So the potential difference drops by exactly the factor $K$. The 16 μF value is a red herring — it never enters the calculation.

Method & Steps

Write the drop in terms of K. $V' = \dfrac{V}{K}$.
Match to the given drop. $\dfrac{V}{K} = \dfrac{V}{8}$.
Solve. $$K = 8$$

Easy Trick

At constant charge, V drops by the same factor that C rises — and that factor is K. "Voltage becomes 1/8th" reads off directly as $K=8$.

Watch the trap: 16 is offered to tempt you into using the capacitance, but $K$ depends only on the ratio of voltages, not on C.

Solution

With charge fixed, $V' = V/K$. Given $V' = V/8$:

$$K = 8$$

b K = 8 dielectric constant