(C) For a deuteron: charge $q_d = e$,mass $m_d = 2m_p = 2m$.For an $\alpha -$ particle: charge $q_{\alpha} = 2e$,mass $m_{\alpha} = 4m_p = 4m$.The for

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If Assertion is correct but Reason is incorrect.

(C) For a deuteron: charge $q_d = e$,mass $m_d = 2m_p = 2m$.For an $\alpha -$ particle: charge $q_{\alpha} = 2e$,mass $m_{\alpha} = 4m_p = 4m$.The force on a charge $q$ in an electric field $E$ is given by $F = qE$.Thus,$F_1 = eE$ and $F_2 = 2eE$. Since $F_1 \neq F_2$,the Reason is incorrect.Now,calculating accelerations using $a = F/m$:$a_1 = F_1 / m_d = (eE) / (2m) = eE / 2m$.$a_2 = F_2 / m_{\alpha} = (2eE) / (4m) = eE / 2m$.Since $a_1 = a_2$,the Assertion is correct.Therefore,the Assertion is correct but the Reason is incorrect.

Class 12 Physics Electric Charges and Fields Motion of Charge particle in Electric filed

Medium

Assertion : $A$ deuteron and an $\alpha -$ particle are placed in an electric field. If $F_1$ and $F_2$ are the forces acting on them and $a_1$ and $a_2$ are their accelerations respectively,then $a_1 = a_2$.
Reason : Forces will be same in an electric field.

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A
If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
B
If both Assertion and Reason are correct,but Reason is not the correct explanation of Assertion.
C
If Assertion is correct but Reason is incorrect.
D
If both the Assertion and Reason are incorrect.

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Electric Charges and Fields

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Motion of Charge particle in Electric filed

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Assertion : $A$ deuteron and an $\alpha -$ particle are placed in an electric field. If $F_1$ and $F_2$ are the forces acting on them and $a_1$ and $a_2$ are their accelerations respectively,then $a_1 = a_2$.Reason : Forces will be same in an electric field.

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Two equal negative charges $-q$ are fixed at the points $(0, a)$ and $(0, -a)$ on the $y-$ axis. $A$ positive charge $Q$ is released from rest at the point $(2a, 0)$ on the $x-$ axis. The charge will

In Millikan's experiment,an oil drop having charge $q$ becomes stationary upon applying a potential difference $V$ between two plates separated by a distance $d$. The weight of the drop is

In Millikan's oil drop experiment,a charged particle is held in equilibrium between two plates by an electric field. If the direction of the electric field is reversed,calculate the acceleration of the charged particle. (in $, g$)

$A$ proton, a deuteron, and an $\alpha$-particle having the same momentum enter a region of uniform electric field between the parallel plates of a capacitor. The electric field is perpendicular to the initial path of the particles. Then the ratio of deflections suffered by them is

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Assertion : $A$ deuteron and an $\alpha -$ particle are placed in an electric field. If $F_1$ and $F_2$ are the forces acting on them and $a_1$ and $a_2$ are their accelerations respectively,then $a_1 = a_2$.
Reason : Forces will be same in an electric field.