Question

What is the magnitude of the electric force between charges of 0.26 C and 0.15 C at a separation of 0.88 m ? k=8.99×109N⋅m2/C2.

Answer 1

Answer:

$F=4.528*10^8$ N

Explanation:

By Coulomb's Law,

$F=k\frac{q_{1} q_{2} }{r^{2} } \\F=\frac{(8.99*10^9)*0.26*0.15}{(0.88)^2}\\ F=4.528*10^8 N$

Ans.

Answer 2

Answer:

The electric force between charges is $4.5\times10^{8} N.$

Explanation:

Given:

r = 0.88m

$q_{2}=0.15C$

$q_{1}=0.26C$

$K=8.99\times 10^{9} Nm^{2} /C^{2}$

Here,

As we all know, The coulomb's constant is denoted by K.

Let the charge of 0.26C is denoted by $q_{1}$.

Let the charge of 0.15C is denoted by $q_{2}$.

The distance between the two charges is denoted by r.

The electric force is denoted by F.

Now,

By the equation,

$F = \frac{Kq_{2}q_{1}}{r^{2} }$

$F=\frac{8.99\times 10^{9} \times0.15\times0.26}{0.88^{2} }\\F=4.5\times10^{8} N$

So, the electric force between charges is $4.5\times10^{8} N.$