Learn basic concept and definition of capacitance, unit of capacitance. Practice formula to find capacitance of a parallel plate capacitor and electric field parallel plate capacitor, equivalent capacitance.
\(Q\propto\Delta V\)
\(Q=C\Delta V\)
\(C=\dfrac{Q}{\Delta V}\)
\(1\,F=1\dfrac{C}{V}\)
\(1\mu F=10^{-6}F\)
\(1p F=10^{-12}F\)
A Doubled
B Half
C Remains the same
D Becomes one-fourth
where \(\sigma=\dfrac{Q}{A}\)
\(E=\dfrac{\sigma}{\epsilon_0} =\dfrac{Q}{\epsilon_0A}\) \(\left[\sigma=\dfrac{Q}{A}\right]\)
\(\Delta V=E.d\) from Gauss Law
\(\Delta V=\dfrac{Qd}{\epsilon_0A}\)
or, \(C=\dfrac{Q(\epsilon_0A)}{Qd}\)
or, \(C=\dfrac{\epsilon_0A}{d}\)
Conclusion
The capacitance of a parallel plate capacitor is proportional to the area of the plates and inversely proportional to the separation between the plates.
A Remains same
B Doubled
C Half
D Becomes one-fourth
_{?\(Q_1=C_1\,\Delta V_c\)}
\(Q_2=C_2\,\Delta V_c\)
\(C_{eq}=\dfrac{Q}{\Delta V_c}\)
or, \(C_{eq}=\dfrac{Q_1+Q_2}{\Delta V_c}\)
or, \(C_{eq}=\dfrac{Q_1}{\Delta V_c}+\dfrac{Q_2}{\Delta V_c}\)
or, \(C_{eq}=C_1+C_2\)
Conclusion :
If two or more capacitors are connected in parallel then their equivalent capacitance is given as
\(C_{eq}=C_1+C_2+C_3+...\)
A \(3\,\mu F\)
B \(5\,\mu F\)
C \(7\,\mu F\)
D \(8\,\mu F\)
Case 1 :
\(Q_{max}=C\mathcal E\) [\(\therefore\)Final charge on the capacitor after connection of battery]
\(\Delta Q =Q_{max}-q\) [\(\therefore\) \(\Delta Q=Q_f-Q_i\)]
\(\Delta Q=C\mathcal E-q\)
Case 2 :
Initially case configuration is as shown in figure.
\(\Delta Q = Q_{max} -(-q)\)
\(\Delta Q=C\mathcal E+q\)
Note- Charge provided by positive terminal of battery is \(C\mathcal E+q\) as battery first levels \(-q\) to zero and then charges the plate upto \(C\mathcal E\).
A \(4\,\mu C\)
B \(3\,\mu C\)
C \(8\,\mu C\)
D \(10\,\mu C\)
Potential difference across the plates = Potential difference across the terminals of battery.
\(C=\dfrac{Q}{\Delta V}\)
\(Q= C\Delta V\)
Note- Whatever be the initial charge, final charge remains the same.
A \(60\,\mu C\)
B \(29\,\mu C\)
C \(49\,\mu C\)
D \(59\,\mu C\)
where \(\sigma=\dfrac{Q}{A}\)
Since distance between plates is small, it can be considered as single conductor.
By Gauss's law,
Flux through Gaussian surface
\(\phi _E=\oint\,E.dA=\dfrac{q_{in}}{\epsilon_0}\)
or, \(\phi _E=2 EA\) ....(1)
Also, \(\phi _E=\dfrac{q_{in}}{\epsilon_0}\) ....(2)
From (1) and (2)
\(2EA=\dfrac{\sigma A}{\epsilon_0}\) \([q_{in}=\sigma\times A]\)
or, \(E_+=\dfrac{\sigma}{2\epsilon_0}\)
\(E_+=\dfrac{Q}{2\epsilon_0A}\)
Similarly for negative charge plate
\(E_-=\dfrac{Q}{2\epsilon_0A}\)
Total field =\(E_++E_-\)
Total field \(=\dfrac{Q}{2\epsilon_0A}+\dfrac{Q}{2\epsilon_0A}=\dfrac{Q}{\epsilon_0A}\)
A \(10^2\,N/C\)
B \(10^5\,N/C\)
C \(10^4\,N/C\)
D \(10^3\,N/C\)