Answer to Question #206241 in Electric Circuits for Ywas

Question

Draw the net electric force on a +1-C charge placed at the center of: 

a. A line segment of length 1.0 m with a +1-C charge and a –1-C charge at its endpoints; 

b. An equilateral triangle of side length 1.0 m with alternating +1-C, –1-C, and +1-C charges 

placed at its vertices;

c. A square of side length 1.0 m with alternating +1-C, –1-C, +1-C, and –1-C charges placed 

at its vertices.

d. Calculate the magnitude and determine the direction of the net electric force on the +1-

C charge placed at the center of configurations in a, b, and c.

Answer 1

Gives

Part

Mid point P(+1C)

Point A(+1C) and B(-1C )

Electric force at centre

$F_{Ap}=\frac{kq_1q_2}{r^2}$

Put value

$F_{AP}=\frac{9\times10^9\times1\times1}{0.5^2}=36\times10^9N$

$$F_{BP}=-\frac{9\times10^9\times1\times1}{0.5^2}=-36\times10^9N$$

$$F_{net}=36\times10^9-36\times10^9=0N$$

Part (b)

PointA (+1C) point B(-1C) pointC(+1c)

$F_{AP}=\frac{kq_1q_2}{r^2_{AP}}$

Put value

$F_{AP}=\frac{9\times10^9\times1\times1}{(\frac{1}{\sqrt3})^2}=27N$

$F_{BP}=-\frac{9\times10^9\times1\times1}{(\frac{1}{\sqrt3})^2}=-27N$

$F_{CP}=\frac{9\times10^9\times1\times1}{(\frac{1}{\sqrt3})^2}=27N$

$F_{net}=F_{AP}+F_{BP}+F_{CP}$

$F_{net}=27-27+27=27N$

Part(c)

Centre of force

$r_{AP}=r_{BP}=r_{CP}=r_{DP}=\frac{1}{\sqrt2}$

$F_{AP}=\frac{KQ_1Q_2}{r^2_Ap}$

Put value

$F_{AP}=\frac{9\times10^9\times1\times1}{(\frac{1}{\sqrt2})^2}=18N$

$F_{BP}=-\frac{9\times10^9\times1\times1}{(\frac{1}{\sqrt2})^2}=-18N$

$F_{CP}=\frac{9\times10^9\times1\times1}{(\frac{1}{\sqrt2})^2}=18N$

$F_{DP}=-\frac{9\times10^9\times1\times1}{(\frac{1}{\sqrt2})^2}=-18N$

$$F_{net}=F_{AP}+F_{BP}+F_{CP}+F_{DP}$$

Put value

$F_{net}=0N$

Part(d)

(a)

Due to centre for Line

F_net=0N

Due to centre of triangle

F_net=27N

Due to square at centre

F_net =0N