(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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$A$ $3 \text{ C}$ charge moves from the point $(0, -2, -5)$ to the point $(5, 1, 2)$ in an electric field expressed as $\vec{E} = 2\hat{i} + 3\hat{j} + 4\hat{k} \text{ N/C}$. The work done in moving the charge is . . . . . . $J$.

DifficultJEE MAIN 2026

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$A$ particle with charge $e$ and mass $m$, moving along the $X$-axis with a uniform speed $u$, enters a region where a uniform electric field $E$ is acting along the $Y$-axis. The particle starts to move in a parabola. Its focal length (neglecting any effect of gravity) is

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$A$ horizontal electric field $E = (mg)/q$ exists as shown in the figure. $A$ mass $m$ with charge $q$ is attached to the end of a light rod of length $l$. If the mass $m$ is released from the position shown in the figure $(\theta = 45^{\circ})$,find the angular velocity of the rod when it passes through the bottom-most position.

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$A$ body having a specific charge of $8\,\mu \text{C/g}$ is resting on a frictionless plane at a distance of $10\,\text{cm}$ from a wall. It starts moving towards the wall when a uniform electric field of $100\,\text{V/m}$ is applied horizontally toward the wall. If the collision of the body with the wall is perfectly elastic,then the time period of the motion will be (in seconds):

DifficultJEE MAIN 2021

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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 $, g$)

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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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$A$ $3 \text{ C}$ charge moves from the point $(0, -2, -5)$ to the point $(5, 1, 2)$ in an electric field expressed as $\vec{E} = 2\hat{i} + 3\hat{j} + 4\hat{k} \text{ N/C}$. The work done in moving the charge is . . . . . . $J$.

$A$ particle with charge $e$ and mass $m$, moving along the $X$-axis with a uniform speed $u$, enters a region where a uniform electric field $E$ is acting along the $Y$-axis. The particle starts to move in a parabola. Its focal length (neglecting any effect of gravity) 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 $, g$)

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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.