Electrostatics
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New answer posted
2 months agoContributor-Level 10
Heat loss; ΔH = U? - U? = 1/2 (C? / (C? +C? ) (V? -V? )²
= 1/2 * (5*2.5)/ (5+2.5) (220-0)²µJ
= 5/ (2*3) * 22*22*100*10? J
= 5*11*22/3 * 10? J = 1210/3 * 10? J = 1210/3 * 10? ³ J * 4 * 10? ²
According to questions
x/100 = 4*10? ²
So, x=4
Note: But given answer by JEE Main x=36
New answer posted
3 months agoContributor-Level 10
Before inserting slab
C_i = ε? A/d
After inserting dielectric slab
C_i = ε? lw/d
C_f = C? + C?
C_f = (Kε? A? /d) + (ε? A? /d)
C_f = 2C_i ⇒ (Kε? wx/d) + (ε? w (l-x)/d) = 2ε? lw/d
4x + l - x = 2l
x = l/3
New answer posted
3 months agoContributor-Level 10
Potential of centre, = V =
Vc = K (Σq)/R
Vc = K (0)/R = 0
Electric field at centre E_B = E_B = ΣE
Let E be electric field produced by each
charge at the centre, then resultant electric field will be
Ec = 0, since equal electric field vectors are acting at equal angle so their resultant is equal to zero.
New answer posted
3 months agoContributor-Level 10
Enet = Eo/k
Enet = E_free - E_bound = qf/Aε? - qb/Aε?
Eo = qf/Aε?
So, (qf-qb)/Aε? = qf/ (kAε? )
qf - qb = qf/k
qb = qf (1 - 1/k)
New answer posted
3 months agoContributor-Level 10
If charge (-Q) at origin is replaced by (+Q), then electric field at the centre of the cube is zero. Thus, electric field at the centre of the cube is as if only (-2Q) charge is present at the origin.
New answer posted
3 months agoContributor-Level 10
When connected in series, equivalent capacitance,
When connected in parallel, equivalent capacitance
C2 = C + C = 2C
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