Question #58243

a balloon with a charge of 6.0 x 10^-5 C is held at a distance of .10m from a second ballon with the same charge what is the repulsive force and why
1

Expert's answer

2016-03-09T08:36:43-0500

Answer on Question 58243, Physics, Electromagnetism

Question:

A balloon with a charge of 6.0105C6.0 \cdot 10^{-5} \, C is held at a distance of 0.10m0.10 \, m from a second balloon with the same charge. What is the repulsive force and why?

Solution:

We can find the magnitude of the repulsive force from the Coulomb's Law:


Frepulsive=kq1q2r2,F_{repulsive} = k \frac{q_1 q_2}{r^2},


here, k=14πε0=9109Nm2C2k = \frac{1}{4\pi\varepsilon_0} = 9 \cdot 10^9 \frac{N \cdot m^2}{C^2} is the Coulomb's constant, q1=q2=6.0105Cq_1 = q_2 = 6.0 \cdot 10^{-5} \, C is the charges of two balloons, rr is the distance between two balloons.

Then, we get:


Frepulsive=9109Nm2C26.0105C6.0105C(0.10m)2=3240N.F_{repulsive} = 9 \cdot 10^9 \frac{N \cdot m^2}{C^2} \cdot \frac{6.0 \cdot 10^{-5} \, C \cdot 6.0 \cdot 10^{-5} \, C}{(0.10 \, m)^2} = 3240 \, N.


Answer:


Frepulsive=3240N.F_{repulsive} = 3240 \, N.


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