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Two circular rings A and B each of radius a = 30 cm are placed coaxially with their axis horizontal in a uniform electric field `E = 10^(5) NC^(-1)` directed vertically upward as shown in figure. Distance between centers of the rings A and B `(C_A and C_B)is 40 cm`. Ring A has positive charge `q_A = 10muC` and B has a negative charge `q_B = -20muC`. A particle of mass m and charge `q = 10muC` is released from rest at the center of ring A. If particle moves along `C_AC_B,` then Work done by electric field, when particle moves from `C_A to C_B` isA. `-1.2 J`B. 1.2 JC. `-3.6 J`D. `3.6J` |
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Answer» Correct Answer - D `W=q(V_(2)-V_(1))=9xx10^(-2)[((10)/(0.3)-(20)/(05))-((10)/(0.5)-(20)/(0.3))]` `=3.6J` Since only conservative forces act on the system, potential energy changes to kinetic energy. `3.6=(1)/(2)mv^(2)` or `v^(2)=72` or `v=6sqrt(2)ms^(-1)` |
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