Question
Derive the formula (/ equation) of capacitance for a paralled plate capacitor.
✓
Answer
→A capacitor made up of two large parallel conducting plates kept at a small distance is called parallel plate capacitor.
→Two parallel conducting plates are arranged parallel to each other as shown in figure. Area of each plate is A and perpendicular distance between the two plates is $d$. Charge on them is + Q and - Q respectively.
→Surface charge density on both the plates is $\sigma\left(=\frac{ Q }{ A }\right)$ and $-\sigma$ respectively.
→ Here the separation (d) between two plates is very small compared to the area of the plates. ( $d^2 \ll<$ A) Therefore, the electric field between the two plates can be considered uniform (So that we can use the formula $E=\frac{\sigma}{2 \varepsilon_0}$ to find out electric field due to both plates.) Electric field in the region above plate I , $E ^{\prime}=\frac{\sigma}{2 \varepsilon_0}-\frac{\sigma}{2 \varepsilon_0}=0$
→Electric field in the region below plate II, $E ^{\prime \prime}=\frac{\sigma}{2 \varepsilon_0}-\frac{\sigma}{2 \varepsilon_0}=0$
→Electric field in the region above plate I,
$E ^{\prime}=\frac{\sigma}{2 \varepsilon_0}-\frac{\sigma}{2 \varepsilon_0}=0$
→Electric field in the region below plate $\Pi$,
$E ^{\prime \prime}=\frac{\sigma}{2 \varepsilon_0}-\frac{\sigma}{2 \varepsilon_0}=0$
→Electric field in the region between two plates,
$\begin{array}{l}
\therefore E =\frac{\sigma}{2 \varepsilon_0}+\frac{\sigma}{2 \varepsilon_0} \\
\therefore E =\frac{\sigma}{\varepsilon_0} \\
\therefore E =\frac{ Q }{\varepsilon_0 A}
\end{array}$
$\left(\because \sigma=\frac{ Q }{ A }\right)$
→Direction of this electric field is from +ve plate to -ve plate.
→The electric field is limited to the region between two plates and is uniform in that entire region.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find the magnetic field B due to a semicircular wire of radius 10.0cm carrying a current of 5.0A at its centre of curvature.
A 10cm long rod carries a charge of $+50\mu\text{C}$ distributed uniformly along its length. Find the magnitude of the electric field at a point 10cm from both the ends of the rod.
→In order to emit the electron from the metallic surface, which physical methods are used to give the minimum energy? Explain.
The maximum tension in the string of an oscillating pendulum is double of the minimum tension. Find the angular amplitude.
Explain giving reasons for the following:
- Photoelectric current in a photocell increases with the increase in the intensity of the incident radiation.
- The stopping potential (V0) varies linearly with the frequency (v) of the incident radiation for a given photosensitive surface with the slope remaining the same for different surfaces.
- Maximum kinetic energy of the photoelectrons is independent of the intensity of incident radiation.
In what way is Gauss’s law in magnetism different from that used in electrostatics? Explain briefly. The Earth’s magnetic field at the Equator is approximately 0.4 G. Estimate the Earth’s magnetic dipole moment. Given: Radius of the Earth = 6400 km.
A capacitor of unknown capacitance, a resistor of 100 $\Omega$ and an inductor of self inductance L = $( 4 /\pi^{2})$henry are connected in series to an ac source of 200V and 50 Hz. Calculate the value of the capacitance and impedance of the circuit when the current is in phase with the voltage. Calculate the power dissipated in the circuit.
→The magnetic field at a point inside a 2.0mH inductor coil becomes 0.80 of its maximum value in 20µs when the inductor is joined to a battery. Find the resistance of the circuit.
A straight wire carrying a current of 12A is bent into a semi-circular arc of radius 2.0cm as shown. What is the magnetic field at O due to:
- Straight segments.
- The semi-circular arc?
Explain the formation of energy band in solids. Draw energy band diagram for (i) a conductor, (ii) an intrinsic semiconductor.