Consider a gravity-free container as shown. T | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The electric potential is $V = y^3 + 2$. The electric field is given by $\vec{E} = -\frac{dV}{dy} \hat{j} = -3y^2 \hat{j}$.At the base $(y=0)$, $V

Vedclass · Accepted Answer

$76$

Vedclass · Answer

(D) The electric potential is $V = y^3 + 2$. The electric field is given by $\vec{E} = -\frac{dV}{dy} \hat{j} = -3y^2 \hat{j}$.At the base $(y=0)$, $V_A = 2 	ext{ V}$. At the top face $(y=2 	ext{ m})$, $V_B = 2^3 + 2 = 10 	ext{ V}$.Using the work-energy theorem: $q(V_A - V_B) = \frac{1}{2}mv^2$. Since $q = -0.5 	ext{ C}$, we have $-0.5(2 - 10) = \frac{1}{2}(2)v^2$, which gives $4 = v^2$, so $v = 2 	ext{ m/s}$ (velocity just before collision).The velocity just before collision is $\vec{v}_i = 2 \hat{j} 	ext{ m/s}$. The velocity just after collision is $\vec{v}_f = -1.5 \hat{j} 	ext{ m/s}$.The change in momentum is $\Delta \vec{p} = m(\vec{v}_f - \vec{v}_i) = 2(-1.5 - 2)\hat{j} = -7 \hat{j} 	ext{ kg m/s}$.The average force exerted by the wall on the ball is $\vec{F}_{wall\_on\_ball} = \frac{\Delta \vec{p}}{\Delta t} = \frac{-7}{0.1} \hat{j} = -70 \hat{j} 	ext{ N}$.By Newton's third law, the force exerted by the ball on the wall is $\vec{F}_{ball\_on\_wall} = +70 \hat{j} 	ext{ N}$.Additionally, the electric field at the top face $(y=2)$ is $\vec{E} = -3(2)^2 \hat{j} = -12 \hat{j} 	ext{ N/C}$.The electric force on the ball is $\vec{F}_e = q\vec{E} = (-0.5)(-12)\hat{j} = 6 \hat{j} 	ext{ N}$.For the container to remain fixed, the external force $\vec{F}_{ext}$ must balance the force from the ball and the electric force on the ball: $\vec{F}_{ext} + \vec{F}_{ball\_on\_wall} + \vec{F}_e = 0$.$\vec{F}_{ext} + 70\hat{j} + 6\hat{j} = 0 \Rightarrow \vec{F}_{ext} = -76 \hat{j} 	ext{ N}$.The magnitude of the required external force is $76 	ext{ N}$.

Similar Questions

The following figures show regular hexagons with charges placed at their vertices. In which of the following cases is the electric field at the center zero?

Six point charges are placed at the vertices of a regular hexagon of side $a$ as shown. If $E$ represents the electric field and $V$ represents the electric potential at the center $O$,then:

$A$ cube of side $b$ has a charge $q$ at each of its vertices. The electric field due to this charge distribution at the centre of this cube will be

Draw the field lines for a bar magnet,a current-carrying finite solenoid,and an electric dipole.

$A$ positively charged thin metal ring of radius $R$ is fixed in the $xy$-plane with its centre at the origin $O$. $A$ negatively charged particle $P$ is released from rest at the point $(0, 0, z_0)$,where $z_0 > 0$. Then the motion of $P$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module