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An infinite number of charge of equal magnitude q, but of opposite sign are placed along the X-axis at x = 1, x = 2, x = 4, x = 8... and so on. The electric potential at the point x = 0 due to these charge will be
- a)q/4πε0
- b)2q/4πε0
- c)14q/4πε0
- d)q /3πε0
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
An infinite number of charge of equal magnitude q, but of opposite sig...
To find the electric potential at x = 0 due to these charges, we can use the formula for electric potential due to a point charge:
V = kq/r
Where:
V is the electric potential,
k is the electrostatic constant (k = 9 x 10^9 Nm^2/C^2),
q is the magnitude of the charge, and
r is the distance from the charge to the point where we want to find the electric potential.
In this case, we have an infinite number of charges with alternating signs along the X-axis. Let's consider the charges at x = 1, x = 2, x = 4, x = 8, and so on.
The distance from each charge to x = 0 is given by their respective positions on the X-axis. So, the distances are 1, 2, 4, 8, and so on.
The magnitude of each charge is the same, q.
Now, let's calculate the electric potential due to each charge separately and then sum them up.
For the charge at x = 1:
V1 = kq/1
For the charge at x = 2:
V2 = k(-q)/2
For the charge at x = 4:
V3 = kq/4
For the charge at x = 8:
V4 = k(-q)/8
And so on.
If we sum up all these individual potentials, we get:
V = V1 + V2 + V3 + V4 + ...
V = kq/1 + k(-q)/2 + kq/4 + k(-q)/8 + ...
Notice that each term is in the form of kq/2^n, where n is the position of the charge on the X-axis.
Using the formula for the sum of an infinite geometric series, we can find the sum of these terms:
V = kq/(1 - (-1/2))
V = kq/(3/2)
V = 2kq/3
Therefore, the electric potential at x = 0 due to these charges is 2kq/3.
V = kq/r
Where:
V is the electric potential,
k is the electrostatic constant (k = 9 x 10^9 Nm^2/C^2),
q is the magnitude of the charge, and
r is the distance from the charge to the point where we want to find the electric potential.
In this case, we have an infinite number of charges with alternating signs along the X-axis. Let's consider the charges at x = 1, x = 2, x = 4, x = 8, and so on.
The distance from each charge to x = 0 is given by their respective positions on the X-axis. So, the distances are 1, 2, 4, 8, and so on.
The magnitude of each charge is the same, q.
Now, let's calculate the electric potential due to each charge separately and then sum them up.
For the charge at x = 1:
V1 = kq/1
For the charge at x = 2:
V2 = k(-q)/2
For the charge at x = 4:
V3 = kq/4
For the charge at x = 8:
V4 = k(-q)/8
And so on.
If we sum up all these individual potentials, we get:
V = V1 + V2 + V3 + V4 + ...
V = kq/1 + k(-q)/2 + kq/4 + k(-q)/8 + ...
Notice that each term is in the form of kq/2^n, where n is the position of the charge on the X-axis.
Using the formula for the sum of an infinite geometric series, we can find the sum of these terms:
V = kq/(1 - (-1/2))
V = kq/(3/2)
V = 2kq/3
Therefore, the electric potential at x = 0 due to these charges is 2kq/3.
Community Answer
An infinite number of charge of equal magnitude q, but of opposite sig...
V=kq/r
we get infinite gp
kq(1+1/2+1/4+.....)
sum of infinite terms a/1-r
we get 2kq
we get infinite gp
kq(1+1/2+1/4+.....)
sum of infinite terms a/1-r
we get 2kq
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An infinite number of charge of equal magnitude q, but of opposite sign are placed along the X-axis at x = 1, x = 2, x = 4, x = 8... and so on. The electric potential at the point x = 0 due to these charge will bea)q/4πε0b)2q/4πε0c)14q/4πε0d)q /3πε0Correct answer is option 'B'. Can you explain this answer? for JEE 2026 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An infinite number of charge of equal magnitude q, but of opposite sign are placed along the X-axis at x = 1, x = 2, x = 4, x = 8... and so on. The electric potential at the point x = 0 due to these charge will bea)q/4πε0b)2q/4πε0c)14q/4πε0d)q /3πε0Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An infinite number of charge of equal magnitude q, but of opposite sign are placed along the X-axis at x = 1, x = 2, x = 4, x = 8... and so on. The electric potential at the point x = 0 due to these charge will bea)q/4πε0b)2q/4πε0c)14q/4πε0d)q /3πε0Correct answer is option 'B'. Can you explain this answer?.
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