
The intensity of electric field required to balance a proton mass $1.7 \times {10^{ - 27}}kg$ and charge $\[1.6{\text{ }} \times {10^{ - 19}}C\]$ is nearly.
A. \[1 \times {10^5}V{m^{ -1}}\]
B. \[1 \times {10^{ - 5}}V{m^{ - 1}}\]
C. \[1 \times {10^7}V{m^{ - 1}}\]
D. \[1 \times {10^{ - 7}}V{m^{ - 1}}\]
Answer
561.9k+ views
Hint:The space around an electric charge in which its influence can be felt is known as the electric field. The intensity of the electric field is given by the ratio of the weight of proton to the charge. So, the magnitude of the electric field is simply defined as the force per charge on the test charge.
Formula used:
Intensity of electric field $ = $Force $ \times $ Charge $ = F \times q$
Complete step by step solution:
Given: Mass of proton $ = 1.7 \times {10^{ - 27}}kg$ and Charge of proton \[ = 1.6{\text{ }} \times {10^{ - 19}}C\]
The space around an electric charge in which its influence can be felt is known as the electric field. The electric field intensity is the measure of the strength of an electric field at any point.The intensity of the electric field is given by the ratio of weight of the proton to the charge and electric field intensity is a vector quantity. It has both magnitude and direction.
Weight of a proton is calculated as the mass of the proton multiplied by gravitational force.
Force ( $F$ ) = Mass of proton ( $m$ ) gravitational force ( $g$ ).
Gravitational force on the surface of earth is: \[9.807{\text{ }}m{s^{ - 2}}\] .
\[F{\text{ }} = {\text{ }}m \times g{\text{ }} = {\text{ }}1.667 \times {10^{ - 26}}{\text{ }}kgm{s^{ - 2}}\]
Therefore, Intensity of electric field ( $E$ ) $ = $Force $ \times $Charge $ = F \times q$
\[\therefore E{\text{ }} = {\text{ }}\dfrac{F}{q} = {\text{ }}\dfrac{{1.667 \times {{10}^{ - 26}}}}
{{1.6 \times {{10}^{ - 19}}}} = {\text{ }}1 \times {10^{ - 7}}V{m^{ - 1}}\]
So,Intensity of electric required to balance a proton \[ = {\text{ }}1 \times {10^{ - 7}}V{m^{ - 1}}\]
Thus, option (D) is correct.
Note:Sometimes students get confused between the two formulas for calculating the intensity of electric field, one formula includes force and charge while the other one include two charges and distance between them.So, use these formulas according to the quantities given in the question.
Formula used:
Intensity of electric field $ = $Force $ \times $ Charge $ = F \times q$
Complete step by step solution:
Given: Mass of proton $ = 1.7 \times {10^{ - 27}}kg$ and Charge of proton \[ = 1.6{\text{ }} \times {10^{ - 19}}C\]
The space around an electric charge in which its influence can be felt is known as the electric field. The electric field intensity is the measure of the strength of an electric field at any point.The intensity of the electric field is given by the ratio of weight of the proton to the charge and electric field intensity is a vector quantity. It has both magnitude and direction.
Weight of a proton is calculated as the mass of the proton multiplied by gravitational force.
Force ( $F$ ) = Mass of proton ( $m$ ) gravitational force ( $g$ ).
Gravitational force on the surface of earth is: \[9.807{\text{ }}m{s^{ - 2}}\] .
\[F{\text{ }} = {\text{ }}m \times g{\text{ }} = {\text{ }}1.667 \times {10^{ - 26}}{\text{ }}kgm{s^{ - 2}}\]
Therefore, Intensity of electric field ( $E$ ) $ = $Force $ \times $Charge $ = F \times q$
\[\therefore E{\text{ }} = {\text{ }}\dfrac{F}{q} = {\text{ }}\dfrac{{1.667 \times {{10}^{ - 26}}}}
{{1.6 \times {{10}^{ - 19}}}} = {\text{ }}1 \times {10^{ - 7}}V{m^{ - 1}}\]
So,Intensity of electric required to balance a proton \[ = {\text{ }}1 \times {10^{ - 7}}V{m^{ - 1}}\]
Thus, option (D) is correct.
Note:Sometimes students get confused between the two formulas for calculating the intensity of electric field, one formula includes force and charge while the other one include two charges and distance between them.So, use these formulas according to the quantities given in the question.
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