Answer to Question #206244 in Electric Circuits for Ywas

Question

2. Draw the net electric field ate 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 field at the center

of each configurations in a, b, an c.

Answer 1

$V_1=\frac{kq}{r_1}$

$r_1=r_2=0.5m$

$V_1=\frac{9\times10^9\times 1}{0.5}=18\times10^{9}V$

$V_2=\frac{kq}{r_2}$

$V_2=-\frac{9\times10^9\times1}{0.5}=-18\times10^{9}V$

Centre of point p

Put Value

$V_{net}=V_1-V_2=0V$

V_net=0V

Part(b)

Potential at centre of triangle

$V_A=\frac{kQ}{r_{AP}}$

$V_A=\frac{9\times10^9 \times1}{\frac{1}{\sqrt3}}=9\sqrt3V$

$V_B=-\frac{9\times10^9 \times1}{\frac{1}{\sqrt3}}=-9\sqrt3V$

$V_c=\frac{9\times10^9 \times1}{\frac{1}{\sqrt3}}=9\sqrt3V$

$V_{net}=V_A+V_B+V_C$

$$V_{net}=9\sqrt3-9\sqrt3+9\sqrt3=9\sqrt3 V$$

Part (c)

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

PointC(+1C) point D(-1C)

Centre point P

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

$V_A=\frac{KQ}{r_{AP}}$

$V_A=\frac{9\times10^9\times1}{\frac{1}{\sqrt2}}=9\sqrt2V$

$V_B=-\frac{9\times10^9\times1}{\frac{1}{\sqrt2}}=-9\sqrt2V$

$V_C=\frac{9\times10^9\times1}{\frac{1}{\sqrt2}}=9\sqrt2V$

$V_D=-\frac{9\times10^9\times1}{\frac{1}{\sqrt2}}=-9\sqrt2V$

$V=V_A+V_B+V_C+V_D$

Put value

$V_{net}=0V$

Part (d)

Electric field

$E_A=\frac{kq}{r^2}$

$$E_A=\frac{9\times10^9\times1}{.5^2}=3.6\times10^{10}N/C$$

$$E_B=-\frac{9\times10^9\times1}{.5^2}=-3.6\times10^{10}N/C$$

$E_{net}=E_A+E_B=0N/C$

Electric field at centre of triangle

$E_A=\frac{kQ}{r^2_{AP}}$

$E_A=\frac{9\times10^9\times1}{(\frac{1}{\sqrt3})^2}=27N/C$

$E_B=-\frac{9\times10^9\times1}{(\frac{1}{\sqrt3})^2}=-27N/C$

$E_C=\frac{9\times10^9\times1}{(\frac{1}{\sqrt3})^2}=27N/C$

Net electric field at centre

$E_{net}=27N/C$

Electric field square at centre

$E_A=\frac{KQ}{r^2_{AP}}$

$E_A=\frac{9\times10^9\times1}{(\frac{1}{\sqrt2})^2}=18N/C$

$E_B=-\frac{9\times10^9\times1}{(\frac{1}{\sqrt2})^2}=-18N/C$

$E_C=\frac{9\times10^9\times1}{(\frac{1}{\sqrt2})^2}=18N/C$

$E_D=-\frac{9\times10^9\times1}{(\frac{1}{\sqrt2})^2}=-18N/C$

Net electric field of center of square

$$E_{net}=E_A+E_B+E_C+E_D=0N/C$$