Question

0.0025 m ОВ 0.0100 m 4. Two point charges are separated by 1.00 x 10-2 m. One charge is -2.8 x 10-8 C; and the other is +2.8 x 10-8 C. The points A and B are located 2.5 x 10-3 m from the lower and upper point charges as shown. 0.0025 m If an electron, which has a charge of 1.60 X 10-19 C, is moved from Test at A to rest at B, what is the change in electric potential energy of the electron? (a) +4.3 X 10-15 J (C) -2.1 x 10-14 J (€) zero joules (b) +5.4 x 10-15 ) (d) -3.2 x 10-14) plates of a parallel plate capacitor each have an area of 0.40 m and are separated by a distance 0.02 m. They are charged until the potential difference between the plates is 3000 V. The charged capacitor is then isolated. Determine the magnitude of the charge on either capacitor plate. (a) 1.8 x 10-7 C (@) 6.8 x 10-7c (b) 2.7 x 10-7c (c) 4.9 x 10-70 (d) 5.4 x 10-7c

Answer 1

899:- 2. = 10.0025m . B en m 0.01 m given, 9, = +2.8 x168c 22 = -2.8x10°c q = Long 2 m 1.60X10-19 when charge q it rest at A Electric potential energy of system atan UA = K241. + K222 1-7 3 kg [option to the • sxvºx1.6ox to"[2.820 mg zoete = -1.07 X1014 5 when charge q is rest of at B Electric potential energy of system up : Kyon + K292 > ke[t ten] • 9xio?21.60216" [pdate the shooting 31.07x1024) Change in potential energy cou. B - - 2.14x1943 -2.1x10195 Any

801":- ,V=3000v LA = 8.40 m2 d=6.02 M Capacitance of parallel plate capacitor charge on plate 0 = CV a AEV - a 0.40x8.85x162 x 3000 0.02. Q- 5.31 x1017 C Approximately D=5.4 x167 cl Correct answer :(d)