Answer the above question.Explian
Attachments:
Answers
Answered by
1
Hi,
The diagram shows three infinitely long uniform line charges placed on the X, Y and Z axis. The work done in moving a unit positive charge from ( 1,1,1 ) to ( 0,1,1 ) is equal to :
The work W done in moving a unit charge from an electric potential V1 to an electric potential V2 is :
where in our case V1 = V(1,1,1) and V2 = V(0,1,1). It is shown here (from a book)... in equation 23-23 that the electric potential a distance R from an infinite, uniform line charge is
V = - 2 k L ln R + constant
where k is Coulomb's constant ( From a book)... and L is the charge per unit length. Let's set the constant equal to zero, since we will ultimately use differences in the electric potential. If we denote the electric potentials due to the line charges at the x, y, and z axes by Vx, Vy, and Vz, respectively, then:
Vx(x,y,z) = - 4 k l ln [sqrt(y^2 + z^2)] = - 2 k l ln (y^2 + z^2)
Vy(x,y,z) = - 6 k l ln [sqrt(x^2 + z^2)] = - 3 k l ln (x^2 + z^2)
Vz(x,y,z) = - 2 k l ln [sqrt(x^2 + y^2)] = - k l ln (x^2 + y^2)
The total electric potential is just the sum of these:
V(x,y,z) = Vx(x,y,z) + Vy(x,y,z) + Vz(x,y,z)
= - k l [ 2 ln (y^2 + z^2) + 3 ln (x^2 + z^2) + ln (x^2 + y^2) ]
Now, the work is
W = V(1,1,1) - V(0,1,1)
= - k l [ 2 ln (2) + 3 ln (2) + ln (2) - 2 ln (2) - 3 ln (1) - ln (1) ]
= - 4 k l ln (2)
= - l ln (2) / ( pi e0 ),
with k = 1 / ( 4 pi e0 ) in MKS units. Unless I made a sign error (in which case the answer is B), the answer seems to be D. Note that this is the work the field does. If they're asking for the work you do to oppose the field, then the answer is B.
Hoping it helps
:-)
The diagram shows three infinitely long uniform line charges placed on the X, Y and Z axis. The work done in moving a unit positive charge from ( 1,1,1 ) to ( 0,1,1 ) is equal to :
The work W done in moving a unit charge from an electric potential V1 to an electric potential V2 is :
where in our case V1 = V(1,1,1) and V2 = V(0,1,1). It is shown here (from a book)... in equation 23-23 that the electric potential a distance R from an infinite, uniform line charge is
V = - 2 k L ln R + constant
where k is Coulomb's constant ( From a book)... and L is the charge per unit length. Let's set the constant equal to zero, since we will ultimately use differences in the electric potential. If we denote the electric potentials due to the line charges at the x, y, and z axes by Vx, Vy, and Vz, respectively, then:
Vx(x,y,z) = - 4 k l ln [sqrt(y^2 + z^2)] = - 2 k l ln (y^2 + z^2)
Vy(x,y,z) = - 6 k l ln [sqrt(x^2 + z^2)] = - 3 k l ln (x^2 + z^2)
Vz(x,y,z) = - 2 k l ln [sqrt(x^2 + y^2)] = - k l ln (x^2 + y^2)
The total electric potential is just the sum of these:
V(x,y,z) = Vx(x,y,z) + Vy(x,y,z) + Vz(x,y,z)
= - k l [ 2 ln (y^2 + z^2) + 3 ln (x^2 + z^2) + ln (x^2 + y^2) ]
Now, the work is
W = V(1,1,1) - V(0,1,1)
= - k l [ 2 ln (2) + 3 ln (2) + ln (2) - 2 ln (2) - 3 ln (1) - ln (1) ]
= - 4 k l ln (2)
= - l ln (2) / ( pi e0 ),
with k = 1 / ( 4 pi e0 ) in MKS units. Unless I made a sign error (in which case the answer is B), the answer seems to be D. Note that this is the work the field does. If they're asking for the work you do to oppose the field, then the answer is B.
Hoping it helps
:-)
Attachments:
yash1160:
Thankyou so much
Similar questions