I\'M CHECKING MY ANSWERS SO PLEASE ANSWER THE QUESTION WITH AN ANSWER AND NOT IN

Question

I'M CHECKING MY ANSWERS SO PLEASE ANSWER THE QUESTION WITH AN ANSWER AND NOT INSTRUCTIONS.

1.) In a right angle traingle ABC, angle ABC = 90 Degree, AB = 2 m, and angle ACB is 41.81 Degree. A point charge of 5*32 nC is placed at point C and another point charge 4* 32 nC is placed at point A. Calculate the electric field at B.

2.) Two charges, one is - 63 nC and other is 2* 63 nC are separated by 2 m distance. What is the electric field at the mid point of the line joining the charges?

3.) Two charges, one is 7 nC and another one is 12 nC are separated by 1 m distance. Calculate distance from the smaller charge where electric field will be zero.

Explanation / Answer

qA = 4*32 nC, qC = 5*32 nC

AB = 2 m

tan(41.81) = AB/AC

AC = 2/tan(41.81) = 2.236 m

E = kq/r^2

EBA = (9*10^9*4*32*10^-9)/(2*2) = 288 N/C

EBC = (9*10^9*5*32*10^-9)/(2.236*2.236) = 288.02 N/C

Angle between EBA and EBC is 90 degrees

electric filed at B is E = [(288)^2+(288.02)^2]^1/2

E = 407.3 N/C

2) q1 = -63 nC, q2 = 2*63 nC, r1 = r2 = 2/2 = 1m

E1 and E2 both are in same direction

E= E1+E2 = [(9*10^9*63*10^9)/(1*1)] + [(9*10^9*2*63*10^9)/(1*1)]

E = 1701 N/C

3) Net electirc field is zero

E1 = E2

kq1/x^2 =kq2/(L-x)^2

7/x^2 = 12/(1-x)^2

x =0.433m from small charge

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