Question #206243

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?


1
Expert's answer
2021-06-14T15:03:43-0400

Gives

(a) r1=r2=0.5mr_1=r_2=0.5m

V1=kqr1V_1=\frac{kq}{r_1}

Put value

V1=9×109×10.5=18×109VV_1=\frac{9\times10^9\times 1}{0.5}=18\times10^{9}V

V2=kqr2V_2=\frac{kq}{r_2}


V1=9×109×10.5=18×109VV_1=-\frac{9\times10^9\times1}{0.5}=-18\times10^{9}V

Net electric potential at symmet of center

Vnet=V1+V2V_{net}=V_1+V_2

Put value


Vnet=18×10918×109=0VV_{net}=18\times10^{9}-18\times10^{9}=0V

(b)

Centre at Point P

Point AputQ1= +1C

Side(a)=1m

Distance BP=CP=AP=13m\frac{1}{\sqrt3} m

Due to +1C centre of potential

VA=kQrAPV_A=\frac{kQ}{r_{AP}}

VA=9×109×113=93VV_A=\frac{9\times10^9 \times1}{\frac{1}{\sqrt3}}=9\sqrt3V

Similarly point B put charge Q2=-1C

VB=9×109×113=93VV_B=-\frac{9\times10^9 \times1}{\frac{1}{\sqrt3}}=-9\sqrt3V

Point c put +1C

Vc=9×109×113=93VV_c=\frac{9\times10^9 \times1}{\frac{1}{\sqrt3}}=9\sqrt3V

Vnet=VA+VB+VCV_{net}=V_A+V_B+V_C


Vnet=9393+93=93VV_{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 =12\frac{1}{\sqrt2} m

VA=KQrAPV_A=\frac{KQ}{r_{AP}}

Put value

VA=9×109×112=92VV_A=\frac{9\times10^9\times1}{\frac{1}{\sqrt2}}=9\sqrt2V

Similarly point B

VB=9×109×112=92VV_B=-\frac{9\times10^9\times1}{\frac{1}{\sqrt2}}=-9\sqrt2V

Similarly point C

VC=9×109×112=92VV_C=\frac{9\times10^9\times1}{\frac{1}{\sqrt2}}=9\sqrt2V

Similarly point D

VD=9×109×112=92VV_D=-\frac{9\times10^9\times1}{\frac{1}{\sqrt2}}=-9\sqrt2V

Net potential at centre

V=VA+VB+VC+VDV=V_A+V_B+V_C+V_D

V=

92V92V+92V92V=0V9\sqrt2V-9\sqrt2V+9\sqrt2V-9\sqrt2V=0V

Part(d)

Electric field

EA=kqr2E_A=\frac{kq}{r^2}

Put value

EA=9×109×1.52=3.6×1010N/CE_A=\frac{9\times10^9\times1}{.5^2}=3.6\times10^{10}N/C


EB=9×109×1.52=3.6×1010N/CE_B=-\frac{9\times10^9\times1}{.5^2}=-3.6\times10^{10}N/C

Enet=EAEB=0E_{net}=E_A-E_B=0

Electric field of triangle

EA=kQrAP2E_A=\frac{kQ}{r^2_{AP}}

EA=9×109×1(13)2=27E_A=\frac{9\times10^9\times1}{(\frac{1}{\sqrt3})^2}=27 N/CEB=9×109×1(13)2=27N/CE_B=-\frac{9\times10^9\times1}{(\frac{1}{\sqrt3})^2}=-27N/C

Ec=9×109×1(13)2=27N/CE_c=\frac{9\times10^9\times1}{(\frac{1}{\sqrt3})^2}=27N/C

Enet=EA+EB+EcE_{net}=E_A+E_B+E_c

Put value

Enet=27N/CE_{net}=27N/C

Electric field square

EA=KQrAP2E_A=\frac{KQ}{r^2_{AP}}

EA=9×109×1(12)2=18N/CE_A=\frac{9\times10^9\times1}{(\frac{1}{\sqrt2})^2}=18N/C

EB=9×109×1(12)2=18N/CE_B=-\frac{9\times10^9\times1}{(\frac{1}{\sqrt2})^2}=-18N/C

EC=9×109×1(12)2=18N/CE_C=\frac{9\times10^9\times1}{(\frac{1}{\sqrt2})^2}=18N/C

ED=9×109×1(12)2=18N/CE_D=-\frac{9\times10^9\times1}{(\frac{1}{\sqrt2})^2}=-18N/C

Enet=EA+EB+EC+EDE_{net}=E_A+E_B+E_C+E_D

Put value

Enet=0N/cE_{net}=0N/c


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS