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fomenos
3 years ago
8

Two point charges of equal magnitude are 7.3 cm apart. At the midpoint of the line connecting them, their combined electric fiel

d has a magnitude of 50 N/C .
Physics
1 answer:
Delicious77 [7]3 years ago
4 0

Answer:

q₁ = q₂ = Q = 14.8 pC

Explanation:

Given that

q₁ = q₂ = Q = ?

Distance between charges = r =7.3 cm = 0.073 m

Combined electric field = E₁ + E₂ = E = 50 N/C

Using formula

E=2\frac{kQ}{r^2}

Rearranging for Q

Q= \frac{Er^2}{2k}\\\\Q=\frac{(50)(.005329)}{2\times 9\times 10^9}

Q=14.8\times 10^{-12}\, C\\\\Q=14.8\, pC

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