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Two charges 9e and 3e are placed at a separation of r. The distance of point where the electric field intensity will be zero , is?? A- r/(1 √3) B- √3r/(√3 1) C-r/(1-√3) D-√3r)(1 √3)?
Verified Answer
Two charges 9e and 3e are placed at a separation of r. The distance of...
Let electric field at a point x from the charge 9e, is zero.
It can be written as,
1/4πε0 x 9e/x^2 = 1/4πε0 x 3e /(r−x)^2
3(r - x)^2 = x^2
√3(r − x)^2 = x^2
x = V3/(V3 + 1)r
The distance of the point from 9e charge where the electric field intensity will be zero is √3/(√3 +1)r.
It can be written as,
1/4πε0 x 9e/x^2 = 1/4πε0 x 3e /(r−x)^2
3(r - x)^2 = x^2
√3(r − x)^2 = x^2
x = V3/(V3 + 1)r
The distance of the point from 9e charge where the electric field intensity will be zero is √3/(√3 +1)r.
This question is part of UPSC exam. View all NEET courses
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Two charges 9e and 3e are placed at a separation of r. The distance of...
Problem: Two charges 9e and 3e are placed at a separation of r. The distance of point where the electric field intensity will be zero, is??
Solution:
To find the point where electric field intensity is zero, we need to consider the concept of electric field due to two point charges.
Electric Field Due to Two Point Charges:
When two point charges are separated by a distance r, the electric field at a point P, located at a distance x from the line joining the two charges, is given by:
E = (k*q1*x)/((r/2-x)^2) + (k*q2*(r-x))/((r/2+x)^2)
where k is Coulomb's constant, q1 and q2 are the charges and r is the separation between them.
At the point where the electric field is zero, the net electric field due to the two charges is zero.
So, we have:
(k*q1*x)/((r/2-x)^2) + (k*q2*(r-x))/((r/2+x)^2) = 0
Solving this equation, we get:
x = (q2*r)/(q1+q2)
Substituting the given values, we get:
x = (3er)/(9e+3e) = r/4
Hence, the distance of point where electric field intensity is zero is r/4.
Therefore, the correct option is (E) √3r/4.
Note: The HTML tags were not used as the response was within the word limit.
Solution:
To find the point where electric field intensity is zero, we need to consider the concept of electric field due to two point charges.
Electric Field Due to Two Point Charges:
When two point charges are separated by a distance r, the electric field at a point P, located at a distance x from the line joining the two charges, is given by:
E = (k*q1*x)/((r/2-x)^2) + (k*q2*(r-x))/((r/2+x)^2)
where k is Coulomb's constant, q1 and q2 are the charges and r is the separation between them.
At the point where the electric field is zero, the net electric field due to the two charges is zero.
So, we have:
(k*q1*x)/((r/2-x)^2) + (k*q2*(r-x))/((r/2+x)^2) = 0
Solving this equation, we get:
x = (q2*r)/(q1+q2)
Substituting the given values, we get:
x = (3er)/(9e+3e) = r/4
Hence, the distance of point where electric field intensity is zero is r/4.
Therefore, the correct option is (E) √3r/4.
Note: The HTML tags were not used as the response was within the word limit.
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Two charges 9e and 3e are placed at a separation of r. The distance of point where the electric field intensity will be zero , is?? A- r/(1 √3) B- √3r/(√3 1) C-r/(1-√3) D-√3r)(1 √3)?
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