Explanation
$$ V_A = \dfrac{1}{4 \pi \epsilon } \dfrac{ q}{r_A} = \dfrac{1}{4 \pi \epsilon } \dfrac{ q}{\sqrt{2+2}}= \dfrac{1}{4 \pi \epsilon } \dfrac{ q}{2}$$
$$ V_B= \dfrac{1}{4 \pi \epsilon } \dfrac{ q}{r_B}= \dfrac{1}{4 \pi \epsilon } \dfrac{ q}{\sqrt{4+0}}= \dfrac{1}{4 \pi \epsilon } \dfrac{ q}{2} $$
$$ V_A - V_B =0 $$
In a charged capacitor the energy is stored in:
When air is replaced by a dielectric medium of constant $$K$$, the capacity of the condenser:
A parallel plate condenser is charged by connecting it to a battery. The battery is disconnected and a glass slab is introduced between the plates. Then,
A metal plate of thickness half the separation between the capacitor plates of capacitance C is inserted. The new capacitance is:
A parallel plate capacitor is first charged and then isolated , and a dielectric slab is introduced between the plates. The quantity that remains unchanged is:
The equivalent capacity between the points ‘A’ and ‘B’ in the following figure will be:
Hint:
If two capacitors with capacitances $${{C}_{1}}\,and\,{{C}_{2}}$$ are connected in parallel then there equivalent capacitance is given as,
$${{C}_{eq.}}={{C}_{1}}\,+\,{{C}_{2}}$$
Step 1: Simplify the Circuit.
The given circuit can be simplified as shown in figure with three capacitors ($${{C}_{1}}={{C}_{2}}={{C}_{3}}=3\mu F$$) in parallel between A and B.
Step 2: Calculate equivalent capacitance between A and B.
Equivalent capacitance between A and B is,
$${{C}_{eq.}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}$$
$$\Rightarrow {{C}_{eq.}}=(3+3+3)\mu F$$
$$\Rightarrow {{C}_{eq.}}=9\mu F$$
Thus, equivalent capacitance between A and B is $$9\mu F$$.
Three capacitors of $$3 \mu F, 2 \mu F $$ and $$6 \mu F$$ are connected in series. When a battery of 10V is connected to this combination then charge on $$3 \mu F$$capacitor will be :
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