Question

UTI 5. (3 points) How far apart are two conducting plates that have an electric field strength of 4.50 x 10 V/m between them, if their potential difference is 15.0 kV?

Answer 1

a) V211K s 6) uz edal E24.5x10²Nlm. Fruit 15x1882 4.8x18xd 112.15 = 3.331 s distance between the plates ca) Work Done onē 20ke 5 00m Eqoda AK-B. OR € 22x1xl«6x16–20: 47 AK B2 0-8 x18x(4-6x16" Dieu. ква { cokes Senergy given to é (b) SkiB z so get the big . un 50x182 1.6x15 to 2 2x1 8 x 66 x d. 2 y For 25x160m it is accelerated 7. 2002 axot i Sat 9om potential will be loov. 2002 qxromatot Y,

8) (a) ve ka RV M V29x10x46x16x10la 2x1514 vz 3.312x106 Potential Energy Uz kq, 92 Uz9x10 x 2 xl 6x10 / 46x1.641819 Excolu 02 3.312x46x16x16x16'y Sleu. U2152-352Mevl potential energy

9) Ford Î E20 -4V.0. TY laveol change in potentiel is Zero'. Vis constant ove-lEode Elds. AV2-SEds Go AV 20.. Change in potential is zero... Vis constant - 11) Pield inside a conductor is zero Baternal electric field is cancelled by internal induced charges ebetrie biedon's Preld becomes zero. Potential Inside conductorz Ka. Ô Va KQ, V22 KB (2) Palle kane kan โ& 261