If an equipotential curve corresponding to a point
charge at origin is given as kx2 + 5y2 = 125, then value
of k is
Answers
Answer:
We can represent electric potentials (voltages) pictorially, just as we drew pictures to illustrate electric fields. Of course, the two are related. Consider Figure 1, which shows an isolated positive point charge and its electric field lines. Electric field lines radiate out from a positive charge and terminate on negative charges. While we use blue arrows to represent the magnitude and direction of the electric field, we use green lines to represent places where the electric potential is constant. These are called equipotential lines in two dimensions, or equipotential surfaces in three dimensions. The term equipotential is also used as a noun, referring to an equipotential line or surface. The potential for a point charge is the same anywhere on an imaginary sphere of radius r surrounding the charge. This is true since the potential for a point charge is given by
V
=
k
Q
r
and, thus, has the same value at any point that is a given distance r from the charge.
Concept:
An equipotential curve is a curve having the same value of the electric potential at every point in an electric field.
The equation of the circle can be defined as
x² + y² = r²
where r is the radius of the circle.
Given:
The equation kx² + 5y² = 125.
Find:
The value of the k in the equation kx² + 5y² = 125.
Solution:
Here, the equipotential curve resembles the equation of the circle.
kx² + 5y² = 125
Dividing by 5 into both sides,
kx²/5 + 5y²/5 = 125/5
kx²/5 + y² = 25
The equation of the circle can be defined as
x² + y² = r²
Compared with the equation of the circle,
k/5 = 1,
k = 5, for the equipotential curve of radius 5.
Hence, the value of k is 5, for the equipotential curve of radius 5.
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