A two point charges 4q and -q are fixed on the x -axis at x=-d2 and x=d2 , respectively. If a third point charge ' q ' is taken from the origin to x=d along the semicircle as shown in the figure, the energy of the charge will:

Option 1 - <p>Decrease by <math><mfrac><mrow><mrow><mn>4</mn><msup><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><mn>3</mn><mi>π</mi><msub><mrow><mrow><mo>∈</mo></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>d</mi></mrow></mrow></mfrac></math></p>
Option 2 - <p>Increase by <math><mfrac><mrow><mrow><mn>2</mn><msup><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><mn>3</mn><mi>π</mi><msub><mrow><mrow><mo>∈</mo></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>d</mi></mrow></mrow></mfrac></math></p>
Option 3 - <p>Increase by <math><mfrac><mrow><mrow><mn>3</mn><msup><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><mn>4</mn><mi>π</mi><msub><mrow><mrow><mo>∈</mo></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>d</mi></mrow></mrow></mfrac></math></p>
Option 4 - <p>&nbsp;Decrease by <math><mfrac><mrow><mrow><msup><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><mn>4</mn><mi>π</mi><msub><mrow><mrow><mo>∈</mo></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>d</mi></mrow></mrow></mfrac></math></p>
3 Views|Posted 6 months ago
Asked by Shiksha User
1 Answer
A
6 months ago
Correct Option - 1
Detailed Solution:

Uinitial =k (4q) (q) (d/2)+k (q) (-q) (d/2)

6kq2dUfinal =4 (4q) (q)3d2+k (q) (-q) (d/2)23kq2dΔU=23-6kq2d-163kq2d

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Similar Questions for you

V = E (1 - e? /τ)
Where τ = RC = 100 × 10? = 10? sec
50V = 100V (1 - e? /10? )
1/2 = 1 - e? ¹?
1/2 = e? ¹?
-ln2 = -10? t
t = ln2/10?
t = 0.693 × 10? sec

 

According to question, we can write

F C A = F C B = K q Q x 2 + ( d 2 ) 2 F = 2 F C A c o s θ = 2 K q Q x [ x 2 + ( d 2 ) 2 ] 3 2

For maxima of force

d F d x = 0 , s o

x = d 2 2

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physics ncert exemplar solutions class 12th chapter two 2025

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