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On a thin rod of length l=1m lying on X axis with one end at origin (0,0) there is a uniformly distributed charge per unit length given by lambda =Kx where K=10^-9Cm2.The work done in displacing a charge q=1000microC from point A0,0.66) to B(0,1) is?
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On a thin rod of length l=1m lying on X axis with one end at origin (0...
Problem: Find the work done in displacing a charge q=1000microC from point A(0,0.66) to B(0,1) on a thin rod of length l=1m lying on X axis with uniformly distributed charge per unit length given by lambda =Kx where K=10^-9Cm2.
Solution:
Step 1: Calculation of Charge Density
The charge density along the rod is given by lambda = Kx, where K = 10^-9 Cm^2.
At x = 0.66m, the charge density is given by:
lambda = Kx = (10^-9)(0.66) = 6.6 x 10^-10 C/m
At x = 1m, the charge density is given by:
lambda = Kx = (10^-9)(1) = 10^-9 C/m
Step 2: Calculation of Electric Field
The electric field at point A(0,0.66) due to the rod is given by:
dE = k(lambda)(dx)/r^2
where k is the Coulomb constant, dx is the length of the element, and r is the distance between the element and the point A.
dE = (9 x 10^9 Nm^2/C^2)(6.6 x 10^-10 C/m)(0.01 m)/(0.66^2 + 0.01)^(3/2)
dE = 2.12 x 10^4 N/C
The electric field at point B(0,1) due to the rod is given by:
dE = k(lambda)(dx)/r^2
dE = (9 x 10^9 Nm^2/C^2)(10^-9 C/m)(0.01 m)/(1^2 + 0.01)^(3/2)
dE = 8.98 x 10^3 N/C
Step 3: Calculation of Work Done
The work done in moving the charge q from point A to B is given by:
W = q(VB - VA)
where VB and VA are the potentials at points B and A, respectively.
VB - VA = -Ed
where d is the distance between points A and B.
d = sqrt[(0 - 0)^2 + (1 - 0.66)^2] = 0.34 m
VB - VA = -(2.12 x 10^4 N/C)(0.34 m)
VB - VA = -7.22 x 10^3 V
W = (1000 x 10^-6 C)(-7.22 x 10^3 V)
W = -7.22 x 10^-3 J
Therefore, the work done in moving the charge q from point A to B is -7.22 x 10^-3 J.
Solution:
Step 1: Calculation of Charge Density
The charge density along the rod is given by lambda = Kx, where K = 10^-9 Cm^2.
At x = 0.66m, the charge density is given by:
lambda = Kx = (10^-9)(0.66) = 6.6 x 10^-10 C/m
At x = 1m, the charge density is given by:
lambda = Kx = (10^-9)(1) = 10^-9 C/m
Step 2: Calculation of Electric Field
The electric field at point A(0,0.66) due to the rod is given by:
dE = k(lambda)(dx)/r^2
where k is the Coulomb constant, dx is the length of the element, and r is the distance between the element and the point A.
dE = (9 x 10^9 Nm^2/C^2)(6.6 x 10^-10 C/m)(0.01 m)/(0.66^2 + 0.01)^(3/2)
dE = 2.12 x 10^4 N/C
The electric field at point B(0,1) due to the rod is given by:
dE = k(lambda)(dx)/r^2
dE = (9 x 10^9 Nm^2/C^2)(10^-9 C/m)(0.01 m)/(1^2 + 0.01)^(3/2)
dE = 8.98 x 10^3 N/C
Step 3: Calculation of Work Done
The work done in moving the charge q from point A to B is given by:
W = q(VB - VA)
where VB and VA are the potentials at points B and A, respectively.
VB - VA = -Ed
where d is the distance between points A and B.
d = sqrt[(0 - 0)^2 + (1 - 0.66)^2] = 0.34 m
VB - VA = -(2.12 x 10^4 N/C)(0.34 m)
VB - VA = -7.22 x 10^3 V
W = (1000 x 10^-6 C)(-7.22 x 10^3 V)
W = -7.22 x 10^-3 J
Therefore, the work done in moving the charge q from point A to B is -7.22 x 10^-3 J.
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On a thin rod of length l=1m lying on X axis with one end at origin (0,0) there is a uniformly distributed charge per unit length given by lambda =Kx where K=10^-9Cm2.The work done in displacing a charge q=1000microC from point A0,0.66) to B(0,1) is?
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