Question

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Problems (20 points each) 1. a) For the following collection of point charges, calculate the magnitude and direction of the total electric field at point P. 750 HC doun to the right 80 cm a The equivele b The total on The power d 30 cm -100 HC 10 cm 100 HC b) What is the electric potential at point P?

Answer 1

Here we apply Coulomb's law and principle of superposition of Electric field and electric potential.

Touc BOCM Soc м 15cm o 30cm Toom1. -loou La J(80)? (15)2 = 78.6 cm VE |(15)2+(102 = 18 cm: HOCA __ = 18 cme - =atlo =78-6 +10 =88.6 cm | Distance of p from 100 uc &-looue us ISCA 4 from '7504 Cdo 88 Gom: from Coulomb's laws Blectric field due to each F = key - 3x68) cloox106) Na 02 20:18)2 -2.78 x 107 NE FPV2 = (gx1600x106) NL_=2.78XIONS (0.18,2 x2 Eg=- x109) (750X160NC (0.886)2 = 8.6 x 10 NIE. See vec for diagram for direction Electric field of positive change is radially

away & negative change is radially Linward to the charge - - --,- E, (see coorfinate axen ISEM fano = lo potom 70 = 33.690 If we add the n Enet = E +ĒJĘ, Eneth = (E. +£) Caso o = 2(2.78 x10 Coej 33:69° Nie = 4.626 X 107 NC Fred), I Eg Sino - E, Sino - Ez =(5-E] Sino - Ez H = So – 86 879 / Tale Magnitude, Enet = | (Enet) + Ener)? = $(4.626)2+(0.8612 Ciond - Enet = 4.71 xlo? NE Direction X = tant § - tan' (-8.640%) Home-loison T or 10.50 below horizontal from th anns (2-626 XoJ

16) Electric Potent is given by *V = kovi V kava , vg ka Neet = vt Ve + Vz (as scaler quantit) p Vevet = kami +212 + terapia #Uned =(9 xloys r 100m - 100m L 18x162 18x162 ned 27.62 x 106 valt. t 750x150 , (0886)