Question

Two point charges, +2.20 μC and -8.00 μC, are separated by 2.60 m. What is the electric potential midway between them? Number Units

Answer 1

Answer:

Electric potential, $V=-4.01\times 10^4\ volts$

Explanation:

Given that,

Charge 1, $q_1=2.2\ \mu C$

Point charge 2, $q_2=-8\ \mu C$

Distance between charges, d = 2.6 m

We need to find the electric potential midway between them. The electric potential is given by :

$V=\dfrac{kq}{r}$

In this case, r = 1.3 m (midway between charges)

$V=\dfrac{kq_1}{r}-\dfrac{kq_2}{r}$

$V=\dfrac{k}{r}(q_1-q_2)$

$V=\dfrac{9\times 10^9}{1.3}(2.2\times 10^{-6}-8\times 10^{-6})$

$V=-40153.84\ volts$

or

$V=-4.01\times 10^4\ volts$

So, the electric potential midway between the charges is $V=-4.01\times 10^4\ volts$. Hence, this is the required solution.