(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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$A$ charged particle,initially at rest at $O$,follows the trajectory shown alongside when released. Such a trajectory is possible in the presence of:

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An inclined plane making an angle of $30^{\circ}$ with the horizontal is placed in a uniform horizontal electric field of $200 \, N/C$ as shown in the figure. $A$ body of mass $1 \, kg$ and charge $5 \, mC$ is allowed to slide down from rest at a height of $1 \, m$. If the coefficient of friction is $0.2$,find the time (in $s$) taken by the body to reach the bottom. $\left[ g = 9.8 \, m/s^2, \sin 30^{\circ} = 0.5, \cos 30^{\circ} = \frac{\sqrt{3}}{2} \approx 0.866 \right]$

DifficultJEE MAIN 2021

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$A$ charged particle of mass $0.003 \, g$ is held stationary in space by placing it in a downward electric field of $6 \times 10^4 \, N/C$. The magnitude of the charge is:

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$A$ charged particle (mass $m$ and charge $q$) moves along the $X$-axis with velocity $V_{0}$. When it passes through the origin,it enters a region having a uniform electric field $\overrightarrow{E} = -E \hat{j}$ which extends up to $x = d$. What is the equation of the path of the electron in the region $x > d$?

DifficultJEE MAIN 2020

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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.

Similar Questions

$A$ simple pendulum of length $L$ is placed between the plates of a parallel plate capacitor having an electric field $E,$ as shown in the figure. Its bob has mass $m$ and charge $q.$ The time period of the pendulum is given by:

$A$ charged particle,initially at rest at $O$,follows the trajectory shown alongside when released. Such a trajectory is possible in the presence of:

$A$ charged particle of mass $0.003 \, g$ is held stationary in space by placing it in a downward electric field of $6 \times 10^4 \, N/C$. The magnitude of the charge 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.