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प्रश्न
There is a uniform spherically symmetric surface charge density at a distance Ro from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V (R(t)) of the distribution as a function of its instantaneous radius R(t) is:
पर्याय
MCQ
रिकाम्या जागा भरा
उत्तर
Explanation:
At any instant t, the total energy of charge distribution is constant.
Uf + Kf = Ui + Ki
i.e., `1/2"mV"^2+("KQ"^2)/(2"R") = 0+("KQ"^2)/(2"R"_0)`
∴ `1/2"mV"^2=("KQ"^2)/(2"R"_0)-("KQ"^2)/(2"R")`
∴ V = `sqrt(2/"m""KQ"^2/2(1/"R"_0-1/"R"))`
∴ V = `sqrt("KQ"^2/"m"(1/"R"_0 -1/"R"))`
= C `sqrt(1/"R"_0-1/"R")`
In addition, the slope of the V - R curve will continue to decrease.
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