Electric Charges and Fields

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New answer posted•a year ago
S
a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- Electric flux through a dipole is always zero, because the positive and negative charges cancel each other out

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New answer posted•a year ago
S
a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- electric field at the axis of the ring is E=KQy/ (R2+z2)3/2 where z is distance .

F=qE=KQqy/ (R2+z2)3/2

When z<

F= 2 π w = -Kz

So force is directly proportional to – distance that is completely defined that is follows S.H.M

(b)w

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New answer posted•a year ago
S
a year ago
This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- two charge -q at A and B

AB=AO+OB=2d and x= small distance perpendicular to O

When x

F=qq/4 1 4 π ε o x 2 ( 10 - 2 ) 2 where AP=BP=r but horizontal components gets cancel out each other and vertical components gets add .

If angle APO=O the net force

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- net electric field at plate γ due to two other plate

From plate 1 ,E1= - Q S 2 ε 0 t o t h e l e f t

From plate 2 ,E2= q S 2 ε 0 t o t h e r i g h t

Total electric field E= E1 + E2 = q - Q S 2 ε 0 to the left , if Q>q

electric field at o due to plate α = - Q S 2 ε 0 t o t h e l e f t

electric field at o due to plate β

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New answer posted•a year ago
S
a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation-

(i) when r

? E . d s   = 1 ε 0 ∫ ρ d V and we know that V = 4 3 π r 3

d V = 3 × 4 3 π r 3 d r = 4 π r 2 d r

  ? E . d s = 1 ε 0 4 π K ∫ r 3 d r

  E(4 ) π r 2 = 4 π K r 4 ε 0 4 = k r 2 4 ε 0 so it is clear that E is radially outwards.

  When r>R

? E . d s = 1 ε 0 ∫ ρ d V

           &nb

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New answer posted•a year ago
New answer posted•a year ago
New question posted•a year ago
New answer posted•a year ago
S
a year ago

1.34 Velocity of the particle, vx = 2.0 ×106m/s

Separation of the two plates, d = 0.5 cm = 0.005 m

Electric field between two plates, E = 9.1 ×102 N/C

Charge of an electron, q = 1.6 ×10-19 C

Mass of an electron, me = 9.1 ×10-31 kg

Let the electron strike the upper plate at the end of the plate L, when d

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