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If potential V = 20/(x2 + y2). The electric field intensity for V is 40(x i + y j)/(x2 + y2)2. State True/False.
- a)True
- b)False
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
If potential V = 20/(x2+ y2). The electric field intensity for V is 40...
Answer: a
Explanation: E = -Grad (V) = -Grad(20/(x2 + y2)) = -(-40x i /(x2 + y2)2 – 40(y j)/(x2 + y2)2) = 40(x i + y j)/(x2 + y2)2. Thus the statement is true.
Explanation: E = -Grad (V) = -Grad(20/(x2 + y2)) = -(-40x i /(x2 + y2)2 – 40(y j)/(x2 + y2)2) = 40(x i + y j)/(x2 + y2)2. Thus the statement is true.
Most Upvoted Answer
If potential V = 20/(x2+ y2). The electric field intensity for V is 40...
Answer:
The given potential is V = 20/(x^2 + y^2).
To find the electric field intensity (E), we need to differentiate the potential function with respect to the position coordinates (x and y).
Differentiating V with respect to x, we get:
∂V/∂x = (∂/∂x)(20/(x^2 + y^2))
= -40x/(x^2 + y^2)^2
Differentiating V with respect to y, we get:
∂V/∂y = (∂/∂y)(20/(x^2 + y^2))
= -40y/(x^2 + y^2)^2
The negative sign indicates that the electric field points in the opposite direction of the corresponding coordinate.
The electric field vector E is given by E = -∇V, where ∇ is the gradient operator.
So, the electric field vector is given by:
E = -∂V/∂x i - ∂V/∂y j
= (-(-40x/(x^2 + y^2)^2)) i - (-40y/(x^2 + y^2)^2) j
= 40x/(x^2 + y^2)^2 i + 40y/(x^2 + y^2)^2 j
= 40(x i + y j)/(x^2 + y^2)^2
Comparing this with the given expression for the electric field intensity, we can see that they are equal. Therefore, the statement is true.
The given potential is V = 20/(x^2 + y^2).
To find the electric field intensity (E), we need to differentiate the potential function with respect to the position coordinates (x and y).
Differentiating V with respect to x, we get:
∂V/∂x = (∂/∂x)(20/(x^2 + y^2))
= -40x/(x^2 + y^2)^2
Differentiating V with respect to y, we get:
∂V/∂y = (∂/∂y)(20/(x^2 + y^2))
= -40y/(x^2 + y^2)^2
The negative sign indicates that the electric field points in the opposite direction of the corresponding coordinate.
The electric field vector E is given by E = -∇V, where ∇ is the gradient operator.
So, the electric field vector is given by:
E = -∂V/∂x i - ∂V/∂y j
= (-(-40x/(x^2 + y^2)^2)) i - (-40y/(x^2 + y^2)^2) j
= 40x/(x^2 + y^2)^2 i + 40y/(x^2 + y^2)^2 j
= 40(x i + y j)/(x^2 + y^2)^2
Comparing this with the given expression for the electric field intensity, we can see that they are equal. Therefore, the statement is true.
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If potential V = 20/(x2+ y2). The electric field intensity for V is 40(x i + y j)/(x2+ y2)2. State True/False.a)Trueb)FalseCorrect answer is option 'A'. Can you explain this answer?
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