Two positive point charges are placed 2m apart. The electric potential at mid point due to these two changes will be:
A. Added to double
B. Reduced to half
C. Remains same (no effect)
D. Cancel each other effect
Right answer is A.
Explanation:
1m 1m
+q--------V=?----------+q
V= k q/r + k q/r
V=k q/r + k q/r
V=2 k q/r.
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Answers
Answer:
A. Added to double
Explanation:
Concept: Potential at any point due to a point charge is given by the expression-
V = q/(4*π*ε0*d)
where q = charge of the point charge (either +ve or -ve)
d = distance of the point from the charge
ε0 = permittivity of free space (8.854187817 × 10⁻¹². F.m⁻¹)
Given: There are two +ve charge with charge +q
d = 1m
To find: The electric potential at mid point
Step by step explanation:
+q--------V--------+q
(1) (2)
V1 due to charge (1) is given by
V1 = q/(4*π*ε0*1)
V1 = q/(4*π*ε0)
V2 due to charge (2) is given by
V2 = q/(4*π*ε0*1)
V2 = q/(4*π*ε0)
Electric potential at mid point (V)
V = V1+V2
V = q/(4*π*ε0) + q/(4*π*ε0)
V= 2q/(4*π*ε0)
Hence, the electric potential at the mid point gets added to double
Answer: A. Added to double
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Answer:
yes A is the correct answer