Two point – like charges Q₁ and Q₂ are positioned at point 1 and 2 respectively. The field intensity to the right of the charge Q₂ on the line that passes through the two charges varies according to a law that is represented schematically in the figure. The field intensity is assumed to be positive if its direction coincides with the positive direction on the x- axis. The distance between the charges is l, then which of the following is true

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

The ratio of the absolute value of the charges $|\frac{Q_{1}}{Q_{2}}| = {(\frac{a + ℓ}{a})}^{2}$

The value of b, where the field intensity is maximum is $b = \frac{ℓ}{{(\frac{ℓ + a}{a})}^{\frac{2}{3}} - 1}$

Q₁ is positive charge while Q₂ is negative charge

The value of b, where the field intensity is maximum is $b = \frac{ℓ}{{(\frac{ℓ + a}{2 a})}^{2 / 3} - 1}$

answer is A, B, D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

At null point, $\frac{Q_{1}}{x} = \frac{Q_{2}}{(L - x)}$ $\frac{Q_{1}}{(L + a_{2})} + \frac{Q_{2}}{a_{2}} = 0 i.e. |Q_{n}| < |Q_{1}|$

charge are opposite as null point is external

$\begin{array}{l} E = \frac{Q_{1}}{(L + a + x)^{2}} - \frac{Q_{2}}{(a + x)^{2}} \\ for E_{max} \Rightarrow \frac{dE}{dx} = 0 \end{array}$

On solving we get

$x = \frac{ℓ}{{(\frac{ℓ + a}{2 a})}^{2 / 3} - 1}$

Watch 3-min video & get full concept clarity