Answer to Question #206243 in Electric Circuits for Ywas

Question

Calculate the total electric potential 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. How did your answers in a, b, and c compare with your electric field calculations for similar 

configurations?

Answer 1

Gives

(a) $r_1=r_2=0.5m$

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

Put value

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

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

$$V_1=-\frac{9\times10^9\times1}{0.5}=-18\times10^{9}V$$

Net electric potential at symmet of center

$V_{net}=V_1+V_2$

Put value

$$V_{net}=18\times10^{9}-18\times10^{9}=0V$$

(b)

Centre at Point P

Point AputQ₁= +1C

Side(a)=1m

Distance BP=CP=AP=$\frac{1}{\sqrt3} m$

Due to +1C centre of potential

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

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

Similarly point B put charge Q₂=-1C

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

Point c put +1C

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

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

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

(C)

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

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

Centre point P

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

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

Put value

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

Similarly point B

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

Similarly point C

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

Similarly point D

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

Net potential at centre

$V=V_A+V_B+V_C+V_D$

V=

$$9\sqrt2V-9\sqrt2V+9\sqrt2V-9\sqrt2V=0V$$

Part(d)

Electric field

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

Put value

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

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

$E_{net}=E_A-E_B=0$

Electric field of triangle

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

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

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

$E_{net}=E_A+E_B+E_c$

Put value

$E_{net}=27N/C$

Electric field square

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

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

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

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

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

$E_{net}=E_A+E_B+E_C+E_D$

Put value

$E_{net}=0N/c$