Question

Determine the electrical force of attraction between two balloons that are charged with the opposite type of charge but the same quantity of charge. The charge on the balloons is 6.0 × 10−7 C and they are separated by a distance of 0.50 m.

Answer 1

Answer:

$1.3 \times 10 {}^{-2} attractive \: (rounded from \: 1.296 \: \times 10 {}^{ - 2} N)$

Explanation:

Step 1: Identify known values in variable form.

Q1 = +6.0 x 10-7 C and Q2 = -6.0 x 10-7 C

d = 0.50 m

Step 2: Identify the requested value

F = ???

Step 3: Substitute and solve

Answer 2

Answer:

The electrical Force of attraction between two balloons will be $-1.6*10^{-22}N$.

Explanation:

Given,

   The Charges on the Balloons are $q_{1}= +6.0*10^{-7}C$ and $q_{2} =-6.0*10^{-7} C$.

They are separated by a distance of 0.50 m.

i.e.                       $r=0.50m$

we need to find the electrical force of attraction between these two balloons.

According to the coulomb's Law, the Electrical Force

                          F = (1/4π∉) $\frac{q_{1}q_{2} }{r^{2} }$

                         $F= \frac{1}{9*10^{9} } \frac{(+6*10^{-7})(-6*10^{-7} ) }{(0.50)^{2} } \\F= \frac{1}{9*10^{9}}\frac{-36*10^{-14} }{0.25 } \\\\F=\frac{-4*10^{-23} }{0.25} \\F=-1.6*10^{-22} N$

Thus, The Electrical Force of Attraction between above Two balloons is $(-1.6*10^{-22} )N$.