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Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

A capacitor is charged by a battery. the battery is removed and another identical uncharged capacitor is connected in parallel. the total electromagnetic energy of resulting system

  1. Decrease by a factor of 2

  2. remains the same

  3. increase by a factor of 2

  4. increase by a factor 4
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Initial energy U1 = 0.5 * C * V^2. When connected to an uncharged identical capacitor, charge Q redistributes such that Q' = Q/2 and V' = V/2. Final energy U2 = 2 * (0.5 * C * (V/2)^2) = 2 * (0.5 * C * V^2 / 4) = 0.5 * (0.5 * C * V^2) = U1 / 2. The energy decreases by a factor of 2.

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

A parallel plate air capacitor has capacity 'C', a distance of separation between plate is 'd' and potential difference 'V' is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is:

  1. $\dfrac{{C{V^2}}}{{2d}}$
  2. $\dfrac{{C{V^2}}}{{d}}$
  3. $\dfrac{{{C^2}{V^2}}}{{2{d^2}}}$
  4. $\dfrac{{{C^2}{V^2}}}{{{d^2}}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The force between plates is F = Q^2 / (2 * e0 * A) = (CV)^2 / (2 * e0 * A). Since C = e0 * A / d, then e0 * A = Cd. Substituting this, F = (C^2 * V^2) / (2 * Cd) = CV^2 / (2d).

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

Two parallel plate capacitor of capacitances C and 2C are conncected in parallel and changed to a potential difference V.If  the bsttery is disconnected and the space between the plate of the capacitor of cpacince c is cpmpletely  filled with a metrial of dielectric constant K, then the potential difference a cross the capacitor will be come

  1. $3V\left( {K + 2} \right)$
  2. $\left( {\frac{{K + 2}}{{3V}}} \right)$
  3. $\left( {\frac{{3V}}{{K + 2}}} \right)$
  4. $\frac{{3\left( {K + 2} \right)}}{V}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Initial charge on C is CV, on 2C is 2CV. Total charge Q = 3CV. When battery is disconnected, Q is constant. New capacitance of first capacitor is KC. Total capacitance C_new = KC + 2C = C(K+2). New potential V_new = Q / C_new = 3CV / (C(K+2)) = 3V / (K+2).

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

A parallel plate capacitor has circular plates of $8.0\ cm$ radius and are separated by $1.0\ mm$. Calculate the capacitance.

  1. $120\ pF$
  2. $140\ pF$
  3. $160\ pF$
  4. $180\ pF$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

C = e0 * A / d = e0 * pi * r^2 / d. C = (8.85e-12 * 3.14 * 0.08^2) / 0.001 = 8.85e-12 * 3.14 * 0.0064 / 0.001 = 1.77e-10 F = 177 pF. The closest option is 180 pF.

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

two similar capacitor are connected to potential v in  parallel order by separating them and joining them in series

  1. the potential on fees plated will be doubled

  2. the charge on the face free plates will increase

  3. the plates in contact will lose their charge

  4. more energy will be stored in the system.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When two capacitors charged to V are connected in parallel, they have charge Q = CV each. If they are disconnected and reconnected in series, the total potential across the combination is V + V = 2V. Thus, the potential on the free plates is doubled.

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

A parallel plate capacitor is made by stacking n equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is 'C' then the resultant capacitance is :

  1. (n+1)C

  2. (n-1)C

  3. nC

  4. C
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here n plates will make $(n-1)$ capacitors with capacitance $C$ and they are in parallel combination.
Thus, the resultant capacitance $=(n-1)C$

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

Two capacitors of capacitance $C _1$ and $C _2$ are connected in parallel across a battery. If $Q _1$ and $Q _2$ respectively be the charges on the capacitors, then $\dfrac {Q _1}{Q _2}$ will be equal to :

  1. $\dfrac {C _2}{C _1}$
  2. $\dfrac {C _1}{C _2}$
  3. $\dfrac {C _1^2}{C _2^2}$
  4. $\dfrac {C _2^2}{C _1^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In parallel combination the both capacitors have same potential , V (say).
So, $Q _1=C _1V$ and $Q _2=C _2V$
$\therefore \dfrac{Q _1}{Q _2}=\dfrac{C _1V}{C _2V}=\dfrac{C _1}{C _2}$