Point charge (q) moves form point (P) to point (S) along the path PQRS as shown in fig. in a uniform electric field E. Pointing along to the positive direction of the x-axis. The co-ordinates of the points P,Q,R and S are (a, b, 0), (2a, 0, 0), (a, –b, 0) and (0, 0, 0) respectively. The work done by the field in the above process is given by the expression - (A) qEa (B) -qEa (C) qEA√2 (D) qE[tex]\sf{\sqrt{\left(2a\right)^2+b^2}}[/tex]. Created by SharmaShivam. Physics

Answer:The Work done is - qEa JoulesExplanation:We know that work done by a conservative force is path independent. Therefore, work done in path PQRS will be equal to the work done in PS path.Now, lets Find the Displacement SP,∴ We got displacement vector.Now, from the formula we know,Here,W Denotes Work done.q Denotes Charge.V Denotes Potential.Now,Now let's write it through Cartesian system.Substituting the values,But as we know that electric field is along x - axis. so electric field along y and z axis is 0.So,∴ The Work done is - qEa Joules.

tex]...

KingdomWarrior

Verified answer

Answer:

The Work done is - qEa Joules

Explanation:

$\rule{300}{1.5}$

We know that work done by a conservative force is path independent. Therefore, work done in path PQRS will be equal to the work done in PS path.

Now, lets Find the Displacement SP,

$\longrightarrow\sf \overrightarrow{\sf SP} =r_{2}-r_{1}\\\\\\\longrightarrow\sf \overrightarrow{\sf SP} = \bigg(0\;\hat{i}+0\;\hat{j}+0\;\hat{k}\bigg)-\bigg(a\;\hat{i}+b\;\hat{j}+0\;\hat{k}\bigg)\\\\\\\longrightarrow\sf \overrightarrow{\sf SP}=-a\;\hat{i}-b\;\hat{j}$

∴ We got displacement vector.

$\rule{300}{1.5}$

Now, from the formula we know,

$\large\bigstar\;\underline{\boxed{\sf W=q\;V}}$

Here,

W Denotes Work done.
q Denotes Charge.
V Denotes Potential.

Now,

$\longrightarrow\sf W=qV\\\\\\\longrightarrow\sf W=q\times \bigg(\overrightarrow{\sf E}\;.\;\overrightarrow{\sf ds}\bigg)\ \ \ \because\Bigg[V=\overrightarrow{\sf E}\;.\;\overrightarrow{\sf ds}\Bigg]$

Now let's write it through Cartesian system.

$\longrightarrow\sf W=q\times \bigg(\overrightarrow{\sf E_{x}}\;.\;\overrightarrow{\sf dx}+\overrightarrow{\sf E_{y}}\;.\;\overrightarrow{\sf dy}+\overrightarrow{\sf E_{z}}\;.\;\overrightarrow{\sf dz}\bigg)$

Substituting the values,

But as we know that electric field is along x - axis. so electric field along y and z axis is 0.So,

$\longrightarrow\sf W=q\times \bigg(E\;\hat{i}\times -a\;\hat{i}+0\;\hat{j}\times -b\;\hat{j}\bigg)\\\\\\\longrightarrow\sf W=q\times \bigg(E\;\hat{i}\times -a\;\hat{i}\bigg)\\\\\\\longrightarrow\sf W=q\times -Ea\\\\\\\longrightarrow\sf W=-\;qEa\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf W=-\;qEa\;J}}}}$

∴ The Work done is - qEa Joules.

$></p><h3><strong>Hence Option - B is correct!</strong></h3><p><img src=$

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