Subject: Electrostatics 1. Consider a dipole consisting of a charge of +q at (1,
ID: 1587060 • Letter: S
Question
Subject: Electrostatics
1. Consider a dipole consisting of a charge of +q at (1,0) cm, and a charge of -q at (-1,0) cm, and recall that the electric potential of a point charge is: V = K (q/r)
(a) Calculate the potential at (0,0)cm. Remember that for potential, you add the contributions from each charge as scalars. Then calculate the potential at (0,1) cm. How about at (0,y) cm?
(b) Calculate the potential at (2,0)cm. Hint: it's not zero.
(c) Calculate the potential at (1,1)cm.
(d) Write down an expression for the potential at (x,y) cm.
Explanation / Answer
a) Vnet = V1 + V2 = kq1/r1^2 + kq2/r2^2 = k/r( q1 + q2) = k/r ( q – q) = 0 V
b) Use below figure to get derivation of Vnet at point (x,y)
Since electric potential obeys superposition principle so potential due to electric dipole as a whole would be sum of potential due to both the charges +q and -q. Thus
Vnet = V1 + V2 = kq1/r1^2 + kq2/r2^2 = (1/40)*(q/r1 – q/r2)
where r1 and r2 respectively are distance of charge +q and -q from point R.
Now draw line PC perpendicular to RO and line QD perpendicular to RO as shown in figure. From triangle POC
cos=OC/OP = OC/a
therefore OC=acos similarly OD=acos
Now ,
r1 = QRRD = OR-OD = r-acos
r2 = PRRC = OR+OC = r+acos
Vnet = (q/40)*[1/( r-acos) - 1/( r+acos)] = (q/40)*[(2acos)/( r2-a2cos2)]
since magnitude of dipole is
|p| = 2qa
Vnet =(1/40)*[(pcos)/( r2-a2cos2)] -(1)
At (x,y) = (2,0)cm , =0 , r= 2+1= 3cm = 0.03m , a= 0.01m
Plugging values in (1),
Vnet =(1/40)*[(pcos)/( r2-a2cos2)] = Vnet =(1/(4*3.14*8.85*10^-12))*[(2*q*0.01*cos0)/( 0.032-0.012cos20)] = q*2.25*10^11
c) At (x,y) = (1.1)cm , =45 , r=sqrt2 = 0.014m , a= 0.01m
Plugging values in (1),
Vnet =(1/40)*[(pcos)/( r2-a2cos2)] = Vnet =(1/(4*3.14*8.85*10^-12))*[(2*q*0.01*cos45)/( 0.0142-0.012cos245)] = q*8.7*10^11
d) From (1),
Vnet =(1/40)*[(pcos)/( r2-a2cos2)]