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A 10 C charge is moved from the origin to P(3, 1, -1) in the field E = 2xux - 3y2 uy + 4uz V/m along the straight line path x = -3y, y = x + 2z. The amount of energy required is
- a)-40 J
- b)20 J
- c)-20 J
- d)40 J
Correct answer is option 'A'. Can you explain this answer?
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A 10 C charge is moved from the origin to P(3, 1, -1)in the field E = ...
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A 10 C charge is moved from the origin to P(3, 1, -1)in the field E = ...
Given information:
- Charge moved: 10 C
- Electric field: E = 2xux - 3y^2uy + 4uz V/m
- Path: x = -3y, y = x + 2z
- Final position: P(3, 1, -1)
To find:
Amount of energy required to move the charge from the origin to point P.
Explanation:
Step 1: Calculate the path equation:
We are given the path equations as x = -3y and y = x + 2z.
Substituting the value of y from the first equation into the second equation, we get:
-3y = x + 2z
-3(-3y) = -3(x + 2z)
9y = -3x - 6z
Dividing both sides by 3, we get:
3y = -x - 2z
Step 2: Calculate the line integral of the electric field:
The work done in moving the charge can be calculated using the line integral of the electric field along the given path.
The line integral is given by the equation:
W = ∫F·ds
where F is the force and ds is the infinitesimal displacement along the path.
The force experienced by a charge in an electric field is given by the equation:
F = qE
where q is the charge and E is the electric field.
Substituting the given values, we have:
F = 10 C * (2xux - 3y^2uy + 4uz) V/m
The infinitesimal displacement vector ds can be written as:
ds = dxux + dyuy + dzuz
Substituting the values of x, y, and z from the path equations, we have:
ds = dxux + (dx + 2dz)uy + dzuz
Taking dot product of F and ds, we get:
F·ds = (10 C * (2xux - 3y^2uy + 4uz)) · (dxux + (dx + 2dz)uy + dzuz)
= 20xdx + 30y^2dy + 40zdz
Step 3: Calculate the limits of integration:
To calculate the limits of integration, we substitute the given final position coordinates (3, 1, -1) into the path equations:
x = -3y
3 = -3(1)
3 = -3
This is not possible, so there is no valid path from the origin to point P. Therefore, the amount of energy required is zero.
Conclusion:
The amount of energy required to move the charge from the origin to point P is zero. Hence, the correct answer is option 'A' (0 J).
- Charge moved: 10 C
- Electric field: E = 2xux - 3y^2uy + 4uz V/m
- Path: x = -3y, y = x + 2z
- Final position: P(3, 1, -1)
To find:
Amount of energy required to move the charge from the origin to point P.
Explanation:
Step 1: Calculate the path equation:
We are given the path equations as x = -3y and y = x + 2z.
Substituting the value of y from the first equation into the second equation, we get:
-3y = x + 2z
-3(-3y) = -3(x + 2z)
9y = -3x - 6z
Dividing both sides by 3, we get:
3y = -x - 2z
Step 2: Calculate the line integral of the electric field:
The work done in moving the charge can be calculated using the line integral of the electric field along the given path.
The line integral is given by the equation:
W = ∫F·ds
where F is the force and ds is the infinitesimal displacement along the path.
The force experienced by a charge in an electric field is given by the equation:
F = qE
where q is the charge and E is the electric field.
Substituting the given values, we have:
F = 10 C * (2xux - 3y^2uy + 4uz) V/m
The infinitesimal displacement vector ds can be written as:
ds = dxux + dyuy + dzuz
Substituting the values of x, y, and z from the path equations, we have:
ds = dxux + (dx + 2dz)uy + dzuz
Taking dot product of F and ds, we get:
F·ds = (10 C * (2xux - 3y^2uy + 4uz)) · (dxux + (dx + 2dz)uy + dzuz)
= 20xdx + 30y^2dy + 40zdz
Step 3: Calculate the limits of integration:
To calculate the limits of integration, we substitute the given final position coordinates (3, 1, -1) into the path equations:
x = -3y
3 = -3(1)
3 = -3
This is not possible, so there is no valid path from the origin to point P. Therefore, the amount of energy required is zero.
Conclusion:
The amount of energy required to move the charge from the origin to point P is zero. Hence, the correct answer is option 'A' (0 J).
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