What are the strength and direction of the electric field at the position indica

Question

What are the strength and direction of the electric field at the position indicated by the dot in the figure(Figure 1)?https://session.masteringphysics.com/problemAsset/1384377/2/27.P28.jpg

Part A: Give your answer in component form. (Assume that x-axis is horisontal and points to the right, and y-axis points upward.) Express your answer in terms of the unit vectors i^ and j^. Express your answer using two significant figures.

PartB: Give your answer as a magnitude and angle measured cw from the positive x-axis. Express your answer using two significant figures.

Part C: CW from the +x-axis Express your answer using two significant figures.

Explanation / Answer

Ans:-

For + 5nC charge, kQ = 8.99*10^9*5*10^-9= 44.95

Electric field strength E = kQ/r^2

Electric field direction is radially away from + charges

At the dot

r1= 2cm = 0.02m

so E1 = 44.95/(0.02)^2 = 112375N/C at direction angle 0

E1 = 112375 i + 0j

E2 = KQ2 / r^2 = 8.99*10^9 *10*10^-9/sqrt (0.02^2 + 0.04^2) = 44950N/C

Angle is 90 – arctan(2/4)= 63.43deg

E2 = 20105.72 i + 40202.77j

E3 = 8.99*10^9*-5*10^-9/(0.04)^2 =- 28093.75N/c

E3 = 0i -28093.75jN/C

So the net field strength at the dot is

E = E1 + E2 + E3 = 132480.72i +12109.02j =133032.96N/C at 5.22deg above the horizontal

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