Two students A and B were taught that electric field near a uniformly charged large surface is normal to the surface and is equal to `(sigma)/(2 in_(0))`. They were also told that field near the surface of a conductor is `(sigma)/(in_(0))` normal to the conductor where s is charge density on the conductor surface. Now both of them were asked to write field between the plates of an ideal parallel plate capacitor having charge density `sigma` and `-sigma` on its plates. Student A said that field can be seen as superposition of field due to two large charged surfaces. He wrote the answer as `E=(sigma)/(2in_(0))+(sigma)/(2in_(0))=(sigma)/(in_(0))` Student B thought that a capacitor has conducting plates and therefore field due to each plate Must be `(sigma)/(in_(0))` . He wrote his answer as `E=(sigma)/(in_(0))+(sigma)/(in_(0))=(2sigma)/(in_(0))` Who is wrong and where is the flaw in thinking ?

Two students A and B were taught that electric field near a uniformly

1.	Two students A and B were taught that electric field near a uniformly charged large surface is normal to the surface and is equal to `(sigma)/(2 in_(0))`. They were also told that field near the surface of a conductor is `(sigma)/(in_(0))` normal to the conductor where s is charge density on the conductor surface. Now both of them were asked to write field between the plates of an ideal parallel plate capacitor having charge density `sigma` and `-sigma` on its plates. Student A said that field can be seen as superposition of field due to two large charged surfaces. He wrote the answer as `E=(sigma)/(2in_(0))+(sigma)/(2in_(0))=(sigma)/(in_(0))` Student B thought that a capacitor has conducting plates and therefore field due to each plate Must be `(sigma)/(in_(0))` . He wrote his answer as `E=(sigma)/(in_(0))+(sigma)/(in_(0))=(2sigma)/(in_(0))` Who is wrong and where is the flaw in thinking ?
Answer» Correct Answer - B is wrong

Discussion

No Comment Found

Related InterviewSolutions

The equivalent capacitance of the circuit shown, between points A and B will be- A. `2/3 muF`B. `5/3 muF`C. `8/3 muF`D. `7/3 muF`
What fraction of the energy drawn from the charging battery is stored in as capacitorA. `100%`B. `75%`C. `50%`D. `25%`
After charging a capacitor the battery is removed. Now by placing a dielectric slab between the platesA. the potential difference between the plates and the energy stored will decreases but the charge on plates will remains sameB. the charge on the plates will decreases and the potential difference between the plates will increasesC. the potential difference between the plates will increases and energy stored will decreases but the charge on the plates will remains sameD. the potential difference, energy stored and the charge will remain unchanged
If a thin metal foil of same area is placed between the two plates of a parallel plate capacitor of capacitance C, then new capacitance will beA. CB. 2CC. 3CD. 4C
Two capacitors `A` and `B` are connected in series with a battery as shown in the figure. When the switch `S` is closed and the two capacitors get charged fully, then A. The potential difference across the plates of A is 4V and across the plates of B is 6VB. The potential difference across the plates of A is 6V and across the plates of B is 4VC. The ratio of electric energies stored in A and B is `2:3`D. The ratio of charges on A and B is 3:2
The capacitance of the capacitors of plate areas `A_1` and `A_2(A_1 lt A_2)` at a distance d is A. `(epsilon_(0)A_(1))/d`B. `(epsilon_(0)A_(2))/d`C. `(epsilon_(0)(A_(1)+A_(2)))/(2d)`D. `(epsilon_(0)sqrt(A_(1)A_(2)))/d`
You have been given a parallel plate air capacitor having capacitance C, a battery of emf `epsilon` and three dielectric blocks having dielectric constants `K_(1), K_(2) and K_(3)` such that `K_(1) gt K_(2) gt K_(3)`. Describe a sequence of steps such as connecting or disconnecting the capacitor to the battery, inserting or taking out of one of the dielectrics etc -so that the capacitor ends up having maximum possible energy stored in it. [Each dielectric block fills completely the space between the plates]. Write this maximum energy.
An air capacitor is charged upto a potential `V_1`. It is connected in parallel to an identical uncharged capacitor filled with a dielectric medium. After redistribution of charge if the potential difference of this combination is V, then the dielectric constant of the substance will beA. `V^(2)(V_(1)-V)`B. `V(V_(1)-V)`C. `V/((V_(1)-V))`D. `((V_(1)-V))/V`
A neutral conducting ball of radius R is connected to one plate of a capacitor (Capacitance = C), the other plate of which is grounded. The capacitor is at a large distance from the ball. Two point charges, q each, begin to approach the ball from infinite distance. The two point charges move in mutually perpendicular directions. Calculate the charge on the capacitor when the two point charges are at distance x and y form the centre of the sphere.
In the circuit shown in the figure all the capacitors have capacitance C. (a) Find the charge on capacitors marked as 1 and 2 when a battery of emf V is connected across points A and B. (b) Find the equivalent capacitance across points C and D marked in the Figure. (c) Find the equivalent capacitance across points E and G.

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment