Physics: Two point charges q1 = +5.70 nc and q2 = −2.30 nc are separated by 45.0 cm. (a) what is the electric potential at a point midway between the charges? 111.3 incorrect: your answer is incorrect. seenkey 136 u

1Explanation:Cu + 2AgNO3 --- Cu(NO3)2 + 2Ag...

Two point charges q1 = +5.70 nc and q2 = −2.30

peno211

11.03.2022

Explanation:

It is given that,

Charge, $Q_1=5.7\ nC=5.7\times 10^{-9}\ C$

Charge, $Q_2=-2.3\ nC=-2.3\times 10^{-9}\ C$

Distance between charges, d = 45 cm = 0.45 m

We need to find the electric potential at a point midway between the charges. It is given by :

$V=\dfrac{kQ_1}{r_1}+\dfrac{kQ_2}{r_2}$

$V=k(\dfrac{Q_1}{r_1}+\dfrac{Q_2}{r_2})$

Midway between the charges means, r = 22.5 cm = 0.225 m

$V=9\times 10^9\times (\dfrac{5.7\times 10^{-9}\ C}{0.225\ m}+\dfrac{-2.3\times 10^{-9}\ C}{0.225\ m})$

$V=136\ volts$

So, the electric potential at a point midway between the charges is 136 volts. Hence, this is the required solution.

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Valfrutos2912

20.02.2022

Explanation:

Cu + 2AgNO3 ---> Cu(NO3)2 + 2Ag

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