CBSE 12th Standard Physics Subject Electrostatic Potential And Capacitance HOT Questions 3 Mark Questions With Solution 2021
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CBSE 12th Standard Physics Subject Electrostatic Potential And Capacitance HOT Questions 3 Mark Questions With Solution 2021
12th Standard CBSE
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Reg.No. :
Physics
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Three capacitors of 1\(\mu F\), 2\(\mu F\) and 3\(\mu F\) are joined in series.
(i) How many times will the capacity become when they are joined in parallel?
(ii) Determine the charge supplied by the battery of 100 V to the maximum resultant capacitor among both the arrangement. -
Two charges +Q and -Q are kept at points (-x2, 0) and (x1, 0) respectively, in the XY-plane. Find the magnitude and direction of the net electric field at the origin (0, 0).
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An electricdipole consists of two opposite charges each of magnitude 1.0\(\times\)10-6C separated by 2cm. The dipole is placed in an external uniform field of 1\(\times\)105 N/C. Find
(i) the maximum torque exterted by the field on the dipole
(ii) the work which an external agent will have to do in turning the dipole through 1800 starting the position, \(\theta\) = 00. -
A charge Q is distributed over two concentric hollow shaperes of radii r and R( > r) such that their surface densities are equal. Find the potential at the common centre.
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(i) How is the electric field due to a charged parallel plate capacitor affected, when a dielectric slab is inserted between the plates fully occupying the intervening region?
(ii) A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has thickness \(\frac{1}{2}\) d, where d is the separation between the plates. Find the expression for the capacitance when the slab is inserted between the plates.
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CBSE 12th Standard Physics Subject Electrostatic Potential And Capacitance HOT Questions 3 Mark Questions With Solution 2021 Answer Keys
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(i) Given, C1 = 1\(\mu F\) C2 = 2\(\mu F\) C3 = 3\(\mu F\)
The combined capacity (Cs) in series combination is given by
\(\frac { 1 }{ { C }_{ s } } =\frac { 1 }{ { C }_{ 1 } } +\frac { 1 }{ { C }_{ 2 } } +\frac { 1 }{ { C }_{ 3 } } =\frac { 1 }{ 1 } +\frac { 1 }{ 2 } +\frac { 1 }{ 3 } =\frac { 11 }{ 6 } \)
\(\Rightarrow { C }_{ s }=\frac { 6 }{ 11 } \mu F\)
The combined capacity (Cp) in parallel combination is given by
Cp = C1 + C2 +C3 = 1 + 2 + 3 = 6\(\mu F\)
\(\Rightarrow { C }_{ p }=11{ C }_{ s }\)
(ii) As, \({ \ C }_{ p }>{ C }_{ s }\)
\(\therefore \) The charge supplied by 100 V battery
\({ q }_{ p }={ C }_{ p }V=6\mu F\times 100=6\times { 10 }^{ -6 }\times 100\)
\(\\ { q }_{ p }=6\times { 10 }^{ -4 }C\)= 600 \(\mu\)C -
Electric field intensity at Point O due to + Q charge,
\({ E }_{ 1 }=\frac { 1 }{ 4\pi { \varepsilon }_{ o } } \times \frac { Q }{ { \left( { x }_{ 2 } \right) }^{ 2 } } (towards \ B) .....(i)\)
Electric field intensity at Point O due to - Q charge,
\({ E }_{ 2 }=\frac { 1 }{ 4\pi { \varepsilon }_{ o } } \times \frac { Q }{ { \left( { x }_{ 1 } \right) }^{ 2 } } (towards \ B).....(i)\)
\({ \because }\) E1 and E2 act along the same direction.
\({ \because }\) Net electric field intensity at point O,
\(E={ E }_{ 1 }+{ E }_{ 2 }=\frac { 1 }{ 4\pi { \varepsilon }_{ o } } \times \frac { Q }{ { \left( { x }_{ 2 } \right) }^{ 2 } } +\frac { 1 }{ 4\pi { \varepsilon }_{ o } } \times \frac { Q }{ { \left( { x }_{ 1 } \right) }^{ 2 } } (towards \ B)\)
\( E=\frac { Q }{ 4\pi { \varepsilon }_{ o } } \left[ \frac { 1 }{ { x }_{ 2 }^{ 2 } } +\frac { 1 }{ { x }_{ 1 }^{ 2 } } \right] \) -
\(Here,q=1\times { 10 }^{ -6 }C,2a=2cm=0.02m\)
\(\therefore p=q\times 2a=(1\times { 10 }^{ -6 })\times 0.02=2\times { 10 }^{ -8 }cm\)
Intensity of the external electric field, E \(=1.0\times { 10 }^{ 5 }N/C\)
\((i){ \tau }_{ max }=pE=(2\times { 10 }^{ -8 })(1.0\times { 10 }^{ 5 })=2\times { 10 }^{ -3 }N-m\ (1)\)
(ii) Net work done in turning the dipole from \(\ 0^0\ to\ 180^0,\)
\( i.e \ W={ \int { _{ { 0 }^{ 0 } }^{ { 180 }^{ 0 } } } }\tau d\theta =\int { _{ { 0 }^{ 0 } }^{ { 180 }^{ 0 } } } pEsin\theta =pE{ [-cos\theta ] }_{ { 0 }^{ 0 } }^{ { 180 }^{ 0 } }\)
\(=-pE(cos{ 180 }^{ 0 }-cos{ 0 }^{ 0 })=2pE\)
\(=2\times (2\times { 10 }^{ -8 })(1\times { 10 }^{ 5 })J=4\times { 10 }^{ -3 }J\ (11/2)\) -
Let q1 and q2 be the charges on them
\(\sigma_1\) = \(\sigma_2\)
\(\therefore \ \frac{q_{1}}{4 \pi r^{2}}=\frac{q_{2}}{4 \pi R^{2}}\)
\(\therefore \ \frac{q_{1}}{q_{2}}=\frac{r^{2}}{R^{2}}\)
i.e. charge on them is distributed in above ratio or \(q_{1}=\frac{r^{2}}{r^{2}+R^{2}} Q \ \text { and } q_{2}=\frac{R^{2}}{r^{2}+R^{2}} Q\)
\(\therefore\) Potential at.centre
V = Potential due to q1 + Potential due to q 2
\(\therefore \ V=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1}}{r}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{2}}{R}=\frac{Q(R+r)}{4 \pi \varepsilon_{0}\left(r^{2}+R^{2}\right)}\) -
The total charge of the capacitor remains conserved on introduction of dielectric slab. Also, the capacitance of capacitor increases to K times of original values.
\(\therefore \\ \) CV = C' V '= (KC) V' \(\Rightarrow V'=\frac { V }{ K } \)
\(\therefore \\ \) New electric field,
\(E'=\frac { V' }{ d } =\left( { \frac { { V }/{ K } }{ d } } \right) =\left( \frac { V }{ d } \right) \frac { 1 }{ k } =\frac { E }{ K } \)
\(\therefore \\ \) On introduction of dielectric medium, new electric field E' becomes \(\frac { 1 }{ K } \) times of its original value (decrease).
