
A charge of 3C moving in a uniform electric field experiences a force of 3000N. The potential between two points situated in the field at a distance 1cm from each other will be
Answer
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Hint: Electric field is the area around every charged body that exerts a force on any other charged body present within its scope. Potential difference is the amount of work done in moving a unit positive test charge from one point to another. Use the mathematical relation between the Electric field intensity, Force and charge to first find the value of the electric field. And later substitute the found value in the expression providing relation of potential difference and electric field to find the required answer.
Formula used:
$F = qE$
$V = Ed$
Complete step-by-step answer:
Electric field is defined as the space around an electric charge in which its influence can be felt. Force experienced by a unit positive charge placed at that point is defined as the electric field intensity at a point. It is a vector quantity and its SI unit is V/m.
Mathematically it can be represented as, $F = qE$
Where F is the force, q is charged and E is denoted as the electric field.
Given, $F = 3000N$, $q = 3C$
$ \Rightarrow E = \dfrac{F}{q} = \dfrac{{3000}}{3} = 1000V/m$
Electric potential is defined as the amount of work done in bringing a unit positive charge from infinity to a specific point against an electric field. Its SI unit is Volt and it is a scalar quantity.
Mathematically it is written as, $V = Ed$
Where V is the potential difference, E is the electric field and d is the distance of point.
According to the question,
Here, $d = 1cm = 0.01m$
$ \Rightarrow V = 1000 \times 0.01 = 10V$
Therefore, the potential between two points situated in the field at a distance 1cm from each other will be10 volts.
Note: Electric field intensity because of a positive charge is always directed away from the charge and the intensity because of a negative charge is always directed towards the charge. The density of an electric field is zero inside a charged hollow conducting hollow sphere. An equipotential surface is defined as a surface on which all points are at the same potential. The electric is taken perpendicular to the equipotential surface and work done in moving a charge on an equipotential surface is zero.
Formula used:
$F = qE$
$V = Ed$
Complete step-by-step answer:
Electric field is defined as the space around an electric charge in which its influence can be felt. Force experienced by a unit positive charge placed at that point is defined as the electric field intensity at a point. It is a vector quantity and its SI unit is V/m.
Mathematically it can be represented as, $F = qE$
Where F is the force, q is charged and E is denoted as the electric field.
Given, $F = 3000N$, $q = 3C$
$ \Rightarrow E = \dfrac{F}{q} = \dfrac{{3000}}{3} = 1000V/m$
Electric potential is defined as the amount of work done in bringing a unit positive charge from infinity to a specific point against an electric field. Its SI unit is Volt and it is a scalar quantity.
Mathematically it is written as, $V = Ed$
Where V is the potential difference, E is the electric field and d is the distance of point.
According to the question,
Here, $d = 1cm = 0.01m$
$ \Rightarrow V = 1000 \times 0.01 = 10V$
Therefore, the potential between two points situated in the field at a distance 1cm from each other will be10 volts.
Note: Electric field intensity because of a positive charge is always directed away from the charge and the intensity because of a negative charge is always directed towards the charge. The density of an electric field is zero inside a charged hollow conducting hollow sphere. An equipotential surface is defined as a surface on which all points are at the same potential. The electric is taken perpendicular to the equipotential surface and work done in moving a charge on an equipotential surface is zero.
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