Capacitance Problems

1. Find the equivalent capacitance of system of capacitors shown below.

Multiple Choice Questions on Capacitors and capacitance for JEE Main and JEE Advanced

Equivalent capacitance of two capacitors each having capacitance C are connected in series.

So,equivalent capacitance = $C/2$

Therefore, the circuit can be drawn like,

$C/2$, $C$and $C/2$are now in parallel. So the equivalent
capacitance,

$C/2$+$C$+ $C/2=2C$

  1. Four capacitors are connected as shown below.

If $C_{1}=2 \mu F$, , $C_{2}=3 \mu F$, $C_{3}=4\mu F$,$
C_{5}=5\mu F$ , calculate the equivalent capacitance between $A$ and
$B$.

From the diagram, we can say that capacitors $C_{1}$ $_{ }$and series combination of $C_{2}$, $C_{3}$ and $C_{4}$ are connected in parallel.

If $C”$ is the equivalent capacitance of $C_{2}$, $C_{3}$ and $C_{4}$, then,

$1/C”=1/3+1/4+1/5$

Or, $1/C”=47/60$

Or, $C”=60/47 \mu F$

Now $C” $and $C_{1}$are in parallel.

So, equivalent capacitance, $C=C_{1}+C”=2+60/47=154/47=3.27 \mu F$

2. A point charge $q$ is located at (2, 4, 3) in xyz coordinate. Find the potential difference between A and B if A = (2, 3, 3) and B = (-2, 3, 3).

We have potential at point A,

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We have potential at point B,

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>V</mi><mi>B</mi></msub><mo>=</mo><mfrac><mi>q</mi><mrow><mn>4</mn><msub><mi>&#x3C0;&#x3B5;</mi><mn>0</mn></msub></mrow></mfrac><mfrac><mn>1</mn><msqrt><msup><mfenced><mrow><msub><mi>x</mi><mn>1</mn></msub><mo>-</mo><msub><mi>x</mi><mn>2</mn></msub></mrow></mfenced><mn>2</mn></msup><mo>+</mo><msup><mfenced><mrow><msub><mi>y</mi><mn>1</mn></msub><mo>-</mo><msub><mi>y</mi><mn>2</mn></msub></mrow></mfenced><mn>2</mn></msup><mo>+</mo><msup><mfenced><mrow><msub><mi>z</mi><mn>1</mn></msub><mo>-</mo><msub><mi>z</mi><mn>2</mn></msub></mrow></mfenced><mn>2</mn></msup></msqrt></mfrac><mspace linebreak="newline"/><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mfrac><mi>q</mi><mrow><mn>4</mn><msub><mi>&#x3C0;&#x3B5;</mi><mn>0</mn></msub></mrow></mfrac><mfrac><mn>1</mn><msqrt><msup><mn>4</mn><mn>2</mn></msup><mo>+</mo><msup><mn>0</mn><mn>2</mn></msup><mo>+</mo><msup><mn>0</mn><mn>2</mn></msup></msqrt></mfrac><mo>=</mo><mfrac><mi>q</mi><mrow><mn>16</mn><msub><mi>&#x3C0;&#x3B5;</mi><mn>0</mn></msub></mrow></mfrac></math>

So,

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>V</mi><mi>A</mi></msub><mo>-</mo><msub><mi>V</mi><mi>B</mi></msub><mo>=</mo><mfrac><mi>q</mi><mrow><mn>4</mn><mi>&#x3C0;</mi><msub><mi>&#x3B5;</mi><mn>0</mn></msub></mrow></mfrac><mo>-</mo><mfrac><mi>q</mi><mrow><mn>16</mn><mi>&#x3C0;</mi><msub><mi>&#x3B5;</mi><mn>0</mn></msub></mrow></mfrac><mspace linebreak="newline"/><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>=</mo><mfrac><mrow><mn>3</mn><mi>q</mi></mrow><mrow><mn>16</mn><mi>&#x3C0;</mi><msub><mi>&#x3B5;</mi><mn>0</mn></msub></mrow></mfrac></math>

3. A parallel plate capacitor is constructed of metal plates, each of area 0.3 $m^{2}.$ If the capacitance is $8nF$, then calculate the plate separation distance.

We have the equation for parallel plate capacitor,

$C=\epsilon _{0}A/d$

Or, $8\times 10^{-9}=8.85\times 10^{-12}(0.3)/d$

Or, $8d=8.85\times 0.3\times 10^{-3}$

Or, $d=3.32\times 10^{-4}$

Or, $d=0.00032 m=0.32 mm$

4. Refer to the below diagram. Find the resulting capacity of a plate capacitor, if the space between the plates of area S is filled with dielectric with permittivity $\epsilon $.

  • We will replace the plate capacitor with two that are parallel. One will be filled with dielectric $l_{1}$ wide, the other will be filled with air and $l-l_{1}$wide.
  • The total capacity is $C=C_{1}+C_{2}$
  • The capacity of a plate capacitor is given by, $C=\epsilon S/d=\epsilon _{0}\epsilon _{r} S/d$
  • Therefore, for capacitor $C_{1}$ and $C_{2}$we get,
  • $C_{1}=\epsilon _{r} S_{1}/d$ ……………….. (1)
  • and $C_{2}=\epsilon _{r} S_{2}/d$ ……………(2)
  • Now, we will determine the areas S1 and S2 using the lengths l, l1 and the total area of the capacitors S.The total area of the capacitor is $S= lx$, where x = width of the capacitor
  • Then, $S_{1}= l_{1}x$ and $S_{2}=(l- l_{1})x$
  • Or, $S_{1}= l_{1}S/l$ and $S_{2}=(l- l_{1})S/l$ (as $S= lx$
  • Now, equation (1) becomes, $C_{1}=\epsilon _{r} l_{1}S/dl$
  • And equation (2) becomes, $C_{2}=\epsilon _{r} (l- l_{1})S/dl$
  • Therefore, the total capacity is $C=C_{1}+C_{2}$
  • <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>C</mi><mo>=</mo><msub><mi>&#x3B5;</mi><mn>0</mn></msub><msub><mi>&#x3B5;</mi><mi>r</mi></msub><mfrac><mrow><msub><mi>l</mi><mn>1</mn></msub><mi>S</mi></mrow><mrow><mi>d</mi><mi>l</mi></mrow></mfrac><mo>+</mo><msub><mi>&#x3B5;</mi><mn>0</mn></msub><mfrac><mrow><mfenced><mrow><mi>l</mi><mo>-</mo><msub><mi>l</mi><mn>1</mn></msub></mrow></mfenced><mi>S</mi></mrow><mrow><mi>d</mi><mi>l</mi></mrow></mfrac><mspace linebreak="newline"/><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mfrac><mrow><msub><mi>&#x3B5;</mi><mn>0</mn></msub><mi>S</mi></mrow><mrow><mi>d</mi><mi>l</mi></mrow></mfrac><mfenced><mrow><msub><mi>&#x3B5;</mi><mi>r</mi></msub><msub><mi>l</mi><mn>1</mn></msub><mo>+</mo><mi>l</mi><mo>-</mo><msub><mi>l</mi><mn>1</mn></msub></mrow></mfenced><mspace linebreak="newline"/><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>=</mo><mfrac><mrow><msub><mi>&#x3B5;</mi><mn>0</mn></msub><mi>S</mi></mrow><mrow><mi>d</mi><mi>l</mi></mrow></mfrac><mfenced open="[" close="]"><mrow><mfenced><mrow><msub><mi>&#x3B5;</mi><mi>r</mi></msub><mo>-</mo><mn>1</mn></mrow></mfenced><msub><mi>l</mi><mn>1</mn></msub><mo>+</mo><mi>l</mi></mrow></mfenced></math>
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