Solve for variable in each figure below
1 answer:
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Answer:
E = 5291.00 N/C
Explanation:
Expression for capacitance is
where
A is area of square plate
D = DISTANCE BETWEEN THE PLATE
We know that capacitrnce and charge is related as
v = 9.523 V
Electric field is given as
=
E = 5291.00 V/m
E = 5291.00 N/C
Answer:
q1 = 7.6uC , -2.3 uC
q2 = 7.6uC , -2.3 uC
( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )
Explanation:
Solution:-
- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.
- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:
Where,
k: The coulomb's constant = 8.99*10^9
- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.
- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.
- Therefore, the force of attraction between the spheres would be:
... Eq 1
- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).
- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,
- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:
.. Eq2
- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows: