Capacitors!

$Q-1$ A capacitor has capacitance $C$ . Is this information sufficient to know what maximum charge the capacitor can contain ? If yes, what is this charge? If no, what other information is needed ?

$Q-2$ The dielectric constant decreases if the temperature is increased . Explain this in terms of polarization of the material.

$Q-3$ When dielectric slab is gradually inserted between the plates of an isolated parallel plate capacitor, the energy of the system decreases. What can you conclude about the force on the slab exerted by the electric field?

$Q-4$ Suppose a charge $+Q_1$ is given to the positive plate and charge $-Q_2$ to the negative plate of a capacitor. What is the "charge on the capacitor"?

#Physics #ElectricityAndMagnetism

Easy Math Editor

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Comments

$Q_1$ No, I think the info is not sufficient at least the potential should be given.

$Q_2$ Dipole constant is the measure of the ease with which an electric dipole to align itself in the direction of electric field. Now, with increase in temperature the kinetic energy of particles constituting the dipole increases and the vibration in the dipole as a whole increases. So, it becomes more difficult for it to align itself with the external field. This results in the reduction of the dipole constant.

$Q_3$ It's got to pull on the slab in the direction of motion, doesn't it? If it's giving up potential energy, it must do positive work...or stash that energy someplace else. (not very sure!)

$Q_4$ By applying Gauss law it can easily be proven that inner faces of plates shall have equal charge of opposite magnitudes, say $q_0$ and $-q_0$ . Field on the outside of plates will be $(Q_1-Q_2)$ / $2\epsilon_0$ and this gives charge of $(Q_1-Q_2)$ / $2$ on the outer plates from Gauss law. This implies inner sides of plates shall have $q_0$ = $(Q_1+Q_2)$ / $2$ and $-q_0$ = $-(Q_1+Q_2)$ / $2$ . Charge on the capacitor hence is $(Q_1+Q_2)$ / $2$ .

Leonardo Jackson - 7 years, 10 months ago

Q_3 : The Electrical Field Energy is decreasing, which shows that the electric field is doing positive work on the dielectric. Therefore we can say that the dielectric is being attracted by the electric field.

shreyam natani - 7 years, 10 months ago

Ans-1 You need Potential Diff. V to calculate max. Charge that can be holded by a capacitor! Q=VC

Ans-2 polarisation is (total dipole)/volume, hemce temperature increases the volume hence decreases the polarisation. Also polarisation = dielectric const.x elect. Field's magnitude! Elect. Field is uniform hence if polariaation decreases, dielect. Constant also decreases.

Ans-3 not sure!

Ans-4 you can not give two seperate charges for the plates in a capacitor! They must be equal in magnitude, opposite in sign! Btw acc. To my readings, the negative charge is induced by the positive hence eq. In mag, oppo in sign. only 95%sure

A Former Brilliant Member - 7 years, 1 month ago

Q1 We need know the electric field for which breakdown of the medium in between the capacitors occours

Achal Sharma - 6 years, 1 month ago

Q_1 : The information is insufficient. One should know the dielectric constant of the medium between the plates of the capacitor so that the medium does not start conducting.

SAMARTH M.O. - 7 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$