
An $ \alpha $ particle is placed in an electric field at a point that has electric potential 5 V. Find its potential energy.
(A) $ 1.6 \times {10^{ - 18}}{\text{J}} $
(B) $ 1.6 \times {10^{ - 14}}{\text{J}} $
(C) $ 1.6 \times {10^{ - 12}}{\text{J}} $
(D) $ 1.6 \times {10^{ - 10}}{\text{J}} $
Answer
553.5k+ views
Hint : The net charge of an alpha particle is twice that of a proton. The voltage or electric potential can be defined as the potential energy possessed by a unit charge in an electric field.
Formula used: In this solution we will be using the following formula;
$\Rightarrow V = \dfrac{{PE}}{q} $ where $ V $ is the electric potential, $ PE $ is the potential energy of the charge, and $ q $ is the charge.
Complete step by step answer
An alpha particle is said to be placed in an electric field at a point where the potential is 5 V. To say that a point in an electric field possesses a particular value of potential, means that one coulomb of charge will possess that value of potential energy.
In essence, the electric potential is defined as the potential possessed by a unit charge in a particular point in an electric field. For example if the electric potential at a point is said to be $ 1V $ , this means that one coulomb of charge will possess a potential energy of 1 J. Hence, the mathematical formula is given by
$\Rightarrow V = \dfrac{{PE}}{q} $ where , $ PE $ is the potential energy of the charge, and $ q $ is the charge.
Hence, for an alpha particle whose charge is $ 2e $ , the potential energy possessed by such charge at a point of electric potential 5 V is
$\Rightarrow PE = V \times q = 5 \times \left( {2 \times 1.6 \times {{10}^{ - 19}}} \right) $
$ \Rightarrow PE = 1.6 \times {10^{ - 18}}J $
Hence, the correct option is A.
Note
Although, in this question, the electric potential was specified at a point, the quantity voltage is more used, which is the difference in electric potential between two points, hence its other name, potential difference. When the electric potential is used, the other point of reference is the point where potential energy is zero. Hence, alternatively, the electric potential at a point can be defined as the work done in bringing a unit charge from infinity (zero potential energy) to that point.
Formula used: In this solution we will be using the following formula;
$\Rightarrow V = \dfrac{{PE}}{q} $ where $ V $ is the electric potential, $ PE $ is the potential energy of the charge, and $ q $ is the charge.
Complete step by step answer
An alpha particle is said to be placed in an electric field at a point where the potential is 5 V. To say that a point in an electric field possesses a particular value of potential, means that one coulomb of charge will possess that value of potential energy.
In essence, the electric potential is defined as the potential possessed by a unit charge in a particular point in an electric field. For example if the electric potential at a point is said to be $ 1V $ , this means that one coulomb of charge will possess a potential energy of 1 J. Hence, the mathematical formula is given by
$\Rightarrow V = \dfrac{{PE}}{q} $ where , $ PE $ is the potential energy of the charge, and $ q $ is the charge.
Hence, for an alpha particle whose charge is $ 2e $ , the potential energy possessed by such charge at a point of electric potential 5 V is
$\Rightarrow PE = V \times q = 5 \times \left( {2 \times 1.6 \times {{10}^{ - 19}}} \right) $
$ \Rightarrow PE = 1.6 \times {10^{ - 18}}J $
Hence, the correct option is A.
Note
Although, in this question, the electric potential was specified at a point, the quantity voltage is more used, which is the difference in electric potential between two points, hence its other name, potential difference. When the electric potential is used, the other point of reference is the point where potential energy is zero. Hence, alternatively, the electric potential at a point can be defined as the work done in bringing a unit charge from infinity (zero potential energy) to that point.
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