Question

4. AM: A 12.0 Volt car battery and an open switch are attached to an empty parallel plate capacitor. Each plate is 0.500 m by 0.500 m and the plate separation is 1.00 cm. (a) Using the appropriate symbols, draw the circuit. (b) Calculate the capacitance of the capacitor in Sl units. (c) The switch is closed. How much charge is stored on the positive plate? (d) Make a detailed sketch of just the capacitor. • Sketch and label the charges on it and the electric field lines. • Calculate the magnitude of the electric field using two methods: AV = E d and E = (1/8) Q/A (e) How many (excess) electrons are on the negative plate? (f) Are there really mobile positive charges on the positive plate? Explain. Can you explain why it takes work to take a neutral capacitor and turn it into a charged capacitor? Hint: charge the negative plate one electron at a time. (h) Calculate the electrical potential energy stored in the capacitor.

Answer 1

Parallel plate capacitor (Set up) Two identical conducting plates are placed parallel to each other. A battery is connected across the plates. When the capacitor is fully charged then the potential drop across the capacitor is equal to the potential of the battery. -Q +Q + E E = Q/A + Wire L Q V Area = A L A parallel-plate capacitor consists of two parallel plates each of area A, separated by a distance L. When the capacitor is charged, the plates carry equal charges of opposite sign. Capacitance can be defined as the capacity to store charge. (EA V, QOV = Q = CV L (Where C is the constant of proportionality called capacitance of the conductor.) The capacitance of a parallel-plate capacitor is proportional to the area of its plates and inversely proportional to the plate separation.

Q = (*2A) v, Q« V =Q = CV (Where C is the constant of proportionality called capacitance of the conductor.) The capacitance of a parallel-plate capacitor is proportional to the area of its plates and inversely proportional to the plate separation. Definition of C, C = Q/V If V = 1 volt, then C = Q(numerically) capacitance of a conductor is numerically equal to the charge that raises its potential by 1 volt.

② Area of the plate= 0.880.5 = 0.25m2 =A and peparation d=100cm The Capacitance - Aco d = 0.25x8.85 x1062 1x102 = 2.21 X10 = 0.22n F Ang' Co Now charge on positive plate=&=cov. - 0.22 x 109x12 = 2.668109c = 2.66 nc Ano method @ 1X102 OV - E xd » E3 OVE 12 1200V/on d method BE- = 2.66X169 AG 0-25X8-85X1012 ~ 1202.25V/m e) Now change on negative plate by quartipation of charge charge on each e= 1.6 X1019 C 2.66X10-9 .= 1.68100 1.6 1079 no ofextraelecto mo

on the no nelle bopitive charges positive plates Battery gireng charge to one plate doca a it gets difficult to charge the difficultas 0 and withdraw charge from the other plate we can pay to-tre plate 9 So battery basto perform some work to more the charges across the charg plaats of the cappelity the electric field will setap between theplater (6) Now energy on the capacity to givenas =&Cv2 Ź XO:22x10 9X1212 15.84nJ Aoro. =