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p1..............................p2 pZ= q(2dx) force acts due to charges +q and -q on any one dipole on the other....here i assume it as p2 Fnet = fq-f-q Fnet = q(E+dE) -q(E-de) where E = 2kp/x3 fnet =...
Electostatics is the study of charges at rest . It involves force, electric field, potential, work done etc for the charges at rest. Moving charges form current & magnetic field. They come under...
The answer is 1Ω Here as the bulb is glowing at full intensity, so voltage across the bulb is = 1.5V Also, for bulb, v = 1.5V, P = 0.45W; Resistance = R = v 2 /P = 1.5*1.5/0.45 = 5Ω So, total...
In the first and the second case we are talking about two individiual charges and thus the field lines between them are exactly like between the two poles of the magnet(which is why they are called...
Gauss Law:- The electric flux ФEthrough any closed surface is equal to the net charge inside the surface,Qinside, divided by ε0:
Gauss Law:- The electric flux ФEthrough any closed surface is equal to the net charge inside the surface,Qinside, divided by ε 0 :
1 st part :you can probably say that it is a symmetrical figure so the tangential opposite components ancel each other . 2 nd part: it is one of the questions from irodov. you can do it as...
-7kp/8 k cap
force on q1 will be= (k.q1.q2)/2a^2
See, the electric field inside the conducting sphere is zero because no charge resides inside its surface. So, by the expression, E = -∂V/∂R V is constant since E is zero. Thus, it will have the same...
p { margin-bottom: 0.25cm; line-height: 120%; }p { margin-bottom: 0.25cm; line-height: 120%; } Dear student, please see the picture below. p { margin-bottom: 0.25cm; line-height: 120%; } Let =...
The gravitational force on the oil drop = mg = 3 x 10 -16 x 10 = 3 x 10 -15 N. Magnitude of the electrostatic force needed balance this force is given by E = mg/10e = (3 x 10 -15 ) / (10 x 1.6 x 10...
p { margin-bottom: 0.25cm; line-height: 120%; } Potential due to a spherical charge distribution at distance r 1 is given by Magnitude of electric field due to spherical charge distribution at...
This question belongs to Mechanics thread. It is answered there. However, I have reproduced the same below: We can use law of conservation of angular momentum and law of conservation of mechanical...
Since, the radius is 1.64 mm i.e. r = 1.64 x 10 -6 m As we know, m=∂*v (where ∂ is density and v is volume) therefore, m = 0.851 x 4/3-pi-r 3 = 1.45 x 10 -17 kg We have already studued that qE= mg...
Since, the charge density of line is 1micro coulomb, i.e. it can be written as 10 -6 C . and radius=1cm =0.01m Now, Electric Field E= ∂/2-pi-epsilon naught-r where, ∂ is linear charge density and r is...
By symmetry net electic field along X-axis at the centre O is zero and electric field along Y-axis will be added up. dE Y =1/4pi-epsilon-naught x dq/r 2 sin@ but dq= ∂(rd@)=q/pi-r x (rd@) = q/pi d@...
p { margin-bottom: 0.25cm; line-height: 120%; } Ooops... there is a small mistake in the diagram. The vector showing the moment of charge –q should point in opposite direction.
correct option is 4.
“... consisting of 2 ncharged charges capacitors ….” is being interpreted as “ consisting of 2 uncharged capacitors” Intial potential difference across 1uF capacitor C1 is V o = 110 V. When the...
ohhhhhhhhhhhhhhhhhhhhhh …. i have solved this type of question ,, by symmetry we take half charge inside and half charge outside .. now solve to get your answer q/2e 0.. thanks
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