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There is one parallel plate capacitor of plate area A and separation between the paltes is d. There is no dielectric placed between the plates of capacitor. The capacitor is connected to a battery of potential difference V volt. While the battery remains connected, some external agent moves the plates apart very slowly to a new separation of 2d. Show and justify that there is no heat loss in the process. |
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Answer» Solution :Initial CAPACITANCE of capacitor `C_(1)=(epsilon_(0)A)/(d)` Amount of energy stored in the capacitor in its initialstate `U_(1)=(1)/(2)((epsilon_(0)A)/(d))V^(2)"" …(i)` LET at ONE instant of time, the separation between the plates be x, then instantaneous capacitancebecomes: `C=(epsilon_(0)A)/(x)` Instantaneous amount of charge stored in the capacitor `Q=CV=(epsilon_(0)A)/(x)V` Electric force between the plates of the capacitor `F_("el") =(Q^(2))/(2epsilon_(0)A)=(((epsilon_(0)A)/(x)V)^(2))/(2epsilon_(0)A)=(1)/(2) (epsilon_(0)AV^(2))/(x^(2))` The same amount of force must be applied by the external agent in opposite direction so that the plates are moved slowly. An external agent applies force in the direction of displacement. Work done by the external agent can be written as follows for infinitesimally small displacement dx. `dW_("ext")=F_(el)dx=(1)/(2) (epsilon_(0)AV^(2))/(x^(2))dx` In the above equation we have used electric force `(F_(el))` because magnitude of force applied by external agent is same as electric force. Total work done by the external agent in moving the plates apart from separation of d to 2d can be calculated by integrating the above equation: `W_("ext") = int_(d)^(2d) (1)/(2) (epsilon_(0)AV^(2))/(x^(2))-dx=(1)/(2)epsilon_(0)AV^(2) int_(d)^(2d) x^(-2)dx` `W_("ext")=(1)/(2)epsilon_(0)AV^(2)[(-1)/(x)]_(d)^(2d)=(1)/(4)((epsilon_(0)A)/(d))V^(2)"" ...(ii)` Capacitance of the capacitor when separation between the plates becomes `2d, C_(2)=(epsilon_(0)A)/(2d)` Energy stored in the capacitor now `U_(2)=(1)/(2)((epsilon_(0)A)/(2d))V^(2)=(1)/(4)((epsilon_(0)A)/(d))V^(2)"" ...(iii)` Initial amount of charge stored in the capacitor `Q_(1)=C_(1)V=(epsilon_(0)A)/(d)V` Final amount of charge stored in the capacitor `Q_(2)=C_(2)V=(epsilon_(0)A)/(2d)V` We can understand that capacitance has decreased and HENCE stored charge has decreased. Charge from the capacitoris transferredto the battery. Work is done on the battery. Work done on the battery `W_(b)=(Q_(1)-Q_(2))V=(1)/(2)((epsilon_(0)A)/(d))V^(2)"" ...(iv)` Work done on the battery means, energy is supplied to the battery. Let H be the amount of heat LOSS in this process then we can write the following equation for energy. `U_(1)+W_("ext") =U_(2)+W_(b)+H""...(v)` We can now substitute from equation (i), (ii), (iii) and (iv) in equation (v) and get the following: `(1)/(2) ((epsilon_(A))/(d))V^(2)+(1)/(4)((epsilon_(A))/(d))V^(2)=(1)/(4)((epsilon_(A))/(d))V^(2)+(1)/(2)((epsilon_(A))/(d))V^(2)+H` `rArr H=0` |
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