This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Right Anastasia, let's go. Step 1: Define the work done to charge a capacitor. When an infinitesimal charge dq is transferred across a potential difference V, the work done dW is given by: dW = V dq Step 2: Relate potential difference to charge and capacitance. The capacitance C of a capacitor is defined as the ratio of the charge q stored on its plates to the potential difference V across them: C = (q)/(V) From this, we can express V as: V = (q)/(C) Step 3: Substitute V into the work done equation and integrate. Substitute the expression for V into the equation for dW: dW = (q)/(C) dq To find the total energy W stored when the capacitor is charged from an initial charge of 0 to a final charge of q, we integrate dW: W = _0^q (q')/(C) dq' Here, q' is used as the integration variable to distinguish it from the upper limit q. Step 4: Perform the integration. W = (1)/(C) _0^q q' dq' W = (1)/(C) [ ((q')^2)/(2) ]_0^q W = (1)/(C) ( (q^2)/(2) - (0^2)/(2) ) W = (q^2)/(2C) This is the first form of the energy stored. Step 5: Derive the other forms using the capacitance definition. We know that q = CV. Substitute this into the expression W = (q^2)/(2C): W = ((CV)^2)/(2C) = (C^2V^2)/(2C) = (1)/(2)CV^2 This is the second form of the energy stored. Now, substitute C = (q)/(V) into the expression W = (1)/(2)CV^2: W = (1)/(2) ((q)/(V)) V^2 = (1)/(2) qV This is the third form of the energy stored. Thus, the electrical energy stored in a capacitor can be expressed in three equivalent forms: W = (1)/(2)CV^2 = (q^2)/(2C) = (1)/(2)qV Send me the next one 📸