A strip of invisible tape 0.14 m long by 0.014 m wide is charged uniformly with

Question

A strip of invisible tape 0.14 m long by 0.014 m wide is charged uniformly with a total net charge of 8 nC (nano le-9) and is suspended horizontally, so it lies along the x axis, with its center at the origin, as shown in the diagram. Calculate the approximate electric field at location m (location A) due to the strip of tape. Do this by dividing the strip into three equal sections, as shown in the diagram, and approximating each section as a point charge what is the approximate electric field at A due to piece #1? 71034.5-103,0xNC what is the approximate electric field at A due to piece #2? &a| |X N/C what is the approximate electric field at A due to piece #37 , What is the approximate net electric field at A? N/C ~! | × N/C

Explanation / Answer

length of each strip = 0.14/3 m = 0.046m

suppose all three strip as point charge.

then distance of 1 from origin.

x = (0.14/6) + (0.14/6) = 0.0934m

and y = 0.03 m

so d^2 = (x^2 + y^2) = 0.0934^2 + 0.03^2

d = 0.098 m

Electric field due to strip 1 = E1

E1 = Kq/r^2

E1 = 9*10^9* 8/3* 10^-9/(0.098^2)

E1 = 2500 N/C

direction tan theta = y/x

tan^-1(0.03/0.0934) = 17.80 deg above x axis.

thus E1 = (E cos 17.8 i + E sin17.8j)

b) for strip 2:

x = 0

y = 0.03 m

E2 = [ (9 * 10^9) * (8/3 * 10^-9) / 0.03^2

E2 = 26667 along y axis .....Ans

E2 = 26667 j N/C)

c)

x = (0.14/6) + (0.14/6) = 0.0934 m

and y = 0.03 m

so d = sqrt(x^2 + y^2) = 0.098 m

E3 = 9*10^9)*(8/3* 10^-9)/0.098^2

E3 = 2500 N/C

tan^-1(0.03/0.098) = 17.8 deg above negative x axis. ....Ans

E3 = - Ecos 17.8 i + E sin 17.8 j)

E3 = -2500 cos 17.8 i + 2500 sin 17.8 j

E3 = -2380 i + 764.23 j

d) Enet = 2Esin(17.8) + E2

= (2 *2500* sin 17.8) + (26667)

Enet = 28195 N/C along y axis.

Enet = 28195 j N/C

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