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An uniform electric field E exists along positive x-axis. The work done in moving a charge 0.5C through a distance 2 m along a direction making an angle 60∘ with x -axis is 10 J. Then what is the magnitude of electric field (in Vm−1)?
- a)15
- b)20
- c)25
- d)30
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
An uniform electric field E exists along positive x-axis. The work don...
Understanding the Problem
To find the magnitude of the electric field (E), we need to analyze the work done (W) in moving a charge (q) through a distance (d) at an angle (θ) to the electric field.
Given Values
- Charge (q) = 0.5 C
- Distance (d) = 2 m
- Angle (θ) = 60 degrees
- Work done (W) = 10 J
Formula for Work Done
The work done in moving a charge in an electric field is given by the formula:
W = q * E * d * cos(θ)
Where:
- W = Work done
- q = Charge
- E = Electric field magnitude
- d = Distance moved
- cos(θ) = Cosine of the angle between the field and the direction of movement
Substituting the Values
Now, substituting the known values into the formula:
10 J = 0.5 C * E * 2 m * cos(60 degrees)
We know that cos(60 degrees) = 0.5.
Thus, the equation simplifies to:
10 = 0.5 * E * 2 * 0.5
This leads to:
10 = 0.5 * E
Solving for E
Now, solving for E:
E = 10 / 0.5 = 20 Vm^-1
Conclusion
The magnitude of the electric field is 20 Vm^-1, which corresponds to option 'B'.
To find the magnitude of the electric field (E), we need to analyze the work done (W) in moving a charge (q) through a distance (d) at an angle (θ) to the electric field.
Given Values
- Charge (q) = 0.5 C
- Distance (d) = 2 m
- Angle (θ) = 60 degrees
- Work done (W) = 10 J
Formula for Work Done
The work done in moving a charge in an electric field is given by the formula:
W = q * E * d * cos(θ)
Where:
- W = Work done
- q = Charge
- E = Electric field magnitude
- d = Distance moved
- cos(θ) = Cosine of the angle between the field and the direction of movement
Substituting the Values
Now, substituting the known values into the formula:
10 J = 0.5 C * E * 2 m * cos(60 degrees)
We know that cos(60 degrees) = 0.5.
Thus, the equation simplifies to:
10 = 0.5 * E * 2 * 0.5
This leads to:
10 = 0.5 * E
Solving for E
Now, solving for E:
E = 10 / 0.5 = 20 Vm^-1
Conclusion
The magnitude of the electric field is 20 Vm^-1, which corresponds to option 'B'.
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An uniform electric field E exists along positive x-axis. The work don...
Force acting on the charged particle due to electric
work done in moving through distance S,
work done in moving through distance S,
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