Answer
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Hint: Here first we will give the relation between electric field and electric potential and find the value of the distance of the point from the charge and then by using the formula for electric potential we will find the magnitude of the charge.
Formula used:
\[E=\dfrac{V}{r}\]
Where, $E$ is the electric field, $V$ is the electric potential and $r$ is the distance of the point from the charge.
$V=\dfrac{kq}{r}$
Where, $k$ is the Boltzmann constant and $q$ is the charge.
Complete step by step answer:
A charge placed in an electric field possesses potential energy and is measured by work done in moving the charge from infinity to that point against the electric field. Electric potential is denoted by $V$. A region around a charged particle or object within which a force would be exerted on other charged particles is known as electric field. The distance between point of charge and center can be calculated by relating electric field and electric potential.
We know that electric field, $E=\dfrac{kq}{{{r}^{2}}}$.
We also know that electric potential, $V=\dfrac{kq}{r}$.
When we divide both the formulas we get,
$r=\dfrac{V}{E}$
$\Rightarrow r=\dfrac{15}{30}$
$\Rightarrow r=0.5m$
So, the distance from the center to the point charge is $0.5\,m$. Now, we will calculate the magnitude of charge by using the formula for electric potential. That is,
$V=\dfrac{kq}{r}$
$\Rightarrow q=\dfrac{Vr}{k}$
$\Rightarrow q=\dfrac{15\times 0.5}{9\times {{10}^{9}}}$
$\therefore q=0.83\times {{10}^{-9}}C=0.83\,nC$
Hence, the value of magnitude of the charge is $0.83\,nC$.
Note: One must not get confused between electric field and electric potential as the basic difference between electric field and electric potential is that electric field is the amount of force per charge while electric potential is the amount of energy or work per charge.
Formula used:
\[E=\dfrac{V}{r}\]
Where, $E$ is the electric field, $V$ is the electric potential and $r$ is the distance of the point from the charge.
$V=\dfrac{kq}{r}$
Where, $k$ is the Boltzmann constant and $q$ is the charge.
Complete step by step answer:
A charge placed in an electric field possesses potential energy and is measured by work done in moving the charge from infinity to that point against the electric field. Electric potential is denoted by $V$. A region around a charged particle or object within which a force would be exerted on other charged particles is known as electric field. The distance between point of charge and center can be calculated by relating electric field and electric potential.
We know that electric field, $E=\dfrac{kq}{{{r}^{2}}}$.
We also know that electric potential, $V=\dfrac{kq}{r}$.
When we divide both the formulas we get,
$r=\dfrac{V}{E}$
$\Rightarrow r=\dfrac{15}{30}$
$\Rightarrow r=0.5m$
So, the distance from the center to the point charge is $0.5\,m$. Now, we will calculate the magnitude of charge by using the formula for electric potential. That is,
$V=\dfrac{kq}{r}$
$\Rightarrow q=\dfrac{Vr}{k}$
$\Rightarrow q=\dfrac{15\times 0.5}{9\times {{10}^{9}}}$
$\therefore q=0.83\times {{10}^{-9}}C=0.83\,nC$
Hence, the value of magnitude of the charge is $0.83\,nC$.
Note: One must not get confused between electric field and electric potential as the basic difference between electric field and electric potential is that electric field is the amount of force per charge while electric potential is the amount of energy or work per charge.
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