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251.

Figure shown a charged conductor resting on an insulating stand. If at the point `P` the charge density is `sigma`, the potential is `V` and the electric field strength is `E`, what are the values of these quantities at point `Q` ? A. `{:("Charge density","Potential","Electric intensity"),(gt sigma,gt V,gt E):}`B. `{:("Charge density","Potential","Electric intensity"),(gt sigma,V,gt E):}`C. `{:("Charge density","Potential","Electric intensity"),(lt sigma,V,E):}`D. `{:("Charge density","Potential","Electric intensity"),(lt sigma,V,lt E):}`

Answer» Correct Answer - D
Discussion
252.

A charge `(-q)` and another charge `(Q)` are kept at two points `A and B` respectively. Keeping the charge `(+Q)` fixed at `B`, the charge `(-q)` at `A` is moved to another point `C` such that `ABC` forms an equilateral triangle of side `l`. The net work done in moving teh charge `(-q)` isA. `(1)/(4pi epsilon_(0))(Qq)/(l)`B. `(1)/(4pi epsilon_(0))(Qq)/(l^(2))`C. `(1)/(4pi epsilon_(0))Qql`D. Zero

Answer» Correct Answer - D
Discussion
253.

A body of capacity `4 muF` is charged to `80V` and another body of capacity `6muF` is charged to `30V`. When they are connected the energy lost by `4 muF` capacitor isA. `7.8 mJ`B. `4.6 mJ`C. `3.2 mJ`D. `2.5 mJ`

Answer» Correct Answer - A
Initial energy of body of capacitance `4 muF` is
`U_(i) = (1)/(2) xx (4 xx 10^(-6))(80)^(2) = 0.0128 J`
Final potential on this body after connection is
`V = (4 xx 80 + 6 xx 30)/(4 + 6) = 50 V1`. So final energy on it
`U_(f) = (1)/(2) xx 4 xx 10^(-6) (50)^(2) = 0.005 J`
Energy lost by this body `= U_(i) - U_(f) = 7.8 mJ`
Discussion
254.

A proton is about `1840` times heavier than an electron. When it is accelerated by a potential difference of `1 kV`. Its kinetic energy will beA. `1840 keV`B. `1//1840 keV`C. `1 keV`D. `920 keV`

Answer» Correct Answer - C
Kinetic energy of proton
`K = eDeltaV = e xx 10^(3)V = 1keV`.
Discussion
255.

Kinetic energy of an electron accelerated in a potential difference of `100 V` isA. `1.6 xx 10^(-17) J`B. `1.6 xx 10^(21) J`C. `1.6 xx 10^(-29) J`D. `1.6 xx 10^(-34) J`

Answer» Correct Answer - A
By using `KE = QV rArr KE = 1.6 xx 10^(-19) xx 100`
`= 1.6 xx 10^(-17) J`
Discussion
256.

An `alpha-` particle is accelerated through a potential difference of `200 V`. The increase in its kinetic energy isA. `100 ev`B. `200 eV`C. `400 eV`D. `800 eV`

Answer» Correct Answer - C
`KE = QV rArr KE = (2e) 200V = 400eV`
Discussion
257.

Three charges `+q,+q and -q` are located at the corners of an equilaterial triangle of side a. Calculate potential energy of the system.

Answer» Correct Answer - `-(q^(2))/(4piepsi_(0)a)`
Discussion
258.

(Figure 3.78) shows three thin concentric spherical shells A, B and C with initial charges on A , B, and C as 3 Q, 2Q, and -Q, respectively. The shells A amd C are connected by a wire such that it does not touch B. Shell B is earthed. Determine the final charges `q_A, q_B, "and" q_C`. .

Answer» The first equation holds for conservation of charge on A and C
`q_A + q_C = 3Q - Q = 2Q` ...(i)
The second equation holds for zero potential of earthed surface
`(V_B)_("surface") + (V_C)_("out") + (V_A)_("in") = 0`
or `(K q_B)/(2) + (K q_C)/(2 R) + (K q_A)/(3 R)= 0`
or `(q_B)/(2) + (q _C)/(2) + (q_A)/(3) = 0` ...(ii)
Also the third equation holds for potential of A and C being equal `V_A = V_C`.
`V_A = (V_A)_("surface") + (V_B)_("out") + (V_C)_("out")`
`V_C = (V_A)_("in") + (V_B)_("in") + (V_C)_("surface")`
`:. (K q _A)/(3 R) + (K q_B)/(3 R) + (k q_C)/(3 R) = (K q_C)/(R) + (K q_B)/(2 R) + (K q _A)/(3 R)`
or `(q_A)/(3) + (q_B)/(3) + (q_C)/(3)= q_C + (q_B)/(2) + (q_A)/(3)` ...(iii)
Now on solving for `q_A, q_B`, and `q_C`, we get
`q_C = Q/(2) , q_A = (3 Q)/(2)`, and `q_B= (-3 Q)/(2)`.
Discussion
259.

The potential produced by a point charge is `V = kQ// r`. Use this information to determine the shape of equipotential surface ?

Answer» `V = kQ //q`, for same distance r from a point, potential is the same. This is possible only for a spherical surface. So shape of equipotential surface should be spherical.
Discussion