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two particles woth each charge magnitude 2.0×10^-7 c but opposite signs are held 15cm apart.what are the magnitude and direction of the electric field E at tge point midway between charges​

Sagot :

onyebuchinnaji

Answer:

The magnitude of the electric field strength is 6.4 x 10⁵ N/C, directed from positive particle to negative particle.

Explanation:

Given;

charge of each particle, Q = 2 x 10⁻⁷ C

distance between the two charges, r = 15 cm = 0.15 m

distance midway between the charges = 0.075 m

The magnitude of the electric field is calculated as;

[tex]E_{net} = E_{+q} + E_{-q}\\\\E_{net} = \frac{kQ}{r_{1/2}^2} + \frac{kQ}{r_{1/2}^2}\\\\E_{net} = 2(\frac{kQ}{r_{1/2}^2})\\\\E_{net} = 2 (\frac{9\times 10^9 \ \times 2\times 10^{-7}}{0.075^2} )\\\\E_{net} = 6.4\times 10^5 \ N/C[/tex]

The direction of the electric field is from positive particle to negative particle.

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