Advertisements
Advertisements
Question
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to `(1/2)` QE, where Q is the charge on the capacitor, and E is the magnitude of the electric field between the plates. Explain the origin of the factor `1/2`.
Advertisements
Solution
Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of Δx.
Hence, work done by the force to do so = FΔx
As a result, the potential energy of the capacitor increases by an amount given as uAΔx.
Where,
u = Energy density
A = Area of each plate
d = Distance between the plates
V = Potential difference across the plates
The work done will be equal to the increase in the potential energy i.e.,
FΔx = uAΔx
`"F" = "uA" = (1/2in_0"E"^2)"A"`
Electric intensity is given by,
`"E" = "V"/"d"`
∴ `"F" = 1/2 in_0("V"/"d")"EA"`
= `1/2(in_0"A" "V"/"d")"E"`
However, capacitance, C = `(in_0"A")/"d"`
∴ `"F" = 1/2("CV")"E"`
Charge on the capacitor is given by,
Q = CV
∴ `F = 1/2"QE"`
The physical origin of the factor, `1/2`, in the force formula lies in the fact that just outside the conductor, the field is E, and inside it is zero.
Hence, it is the average value, `"E"/2`, of the field that contributes to the force.
RELATED QUESTIONS
Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere's circuital law to include the term due to displacement current.
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm−1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?
Define the capacitance of a capacitor. Obtain the expression for the capacitance of a parallel plate capacitor in vacuum in terms of plate area A and separation d between the plates.
A ray of light falls on a transparent sphere with centre C as shown in the figure. The ray emerges from the sphere parallel to the line AB. Find the angle of refraction at A if the refractive index of the material of the sphere is \[\sqrt{3}\].
A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.
A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/3, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.
A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness 2d/3, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.
Define the capacitance of a capacitor and its SI unit.
A parallel-plate capacitor has plate area 20 cm2, plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 kΩ resistor. Find the energy of the capacitor 8.9 μs after the connections are made.
A parallel-plate capacitor is filled with a dielectric material of resistivity ρ and dielectric constant K. The capacitor is charged and disconnected from the charging source. The capacitor is slowly discharged through the dielectric. Show that the time constant of the discharge is independent of all geometrical parameters like the plate area or separation between the plates. Find this time constant.
Answer the following question.
Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.
Solve the following question.
A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from the battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor.
For a one dimensional electric field, the correct relation of E and potential V is _________.
In a parallel plate capacitor, the capacity increases if ______.
Two identical capacitors are joined in parallel, charged to a potential V, separated and then connected in series, the positive plate of one is connected to the negative of the other. Which of the following is true?
A parallel plate capacitor is charged by connecting it to a battery through a resistor. If I is the current in the circuit, then in the gap between the plates:
