Capacitors with Dielectric
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359233 There is an air filled 1 \(pF\) parallel plate capacitor. When the plate separation is doubled and the space is filled with wax, the capacitance increases to 2 \(pF\). The dielectric constant of wax is

1 \(4\)
2 \(2\)
3 \(8\)
4 \(6\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359234 Two dielectric slabs of constant \({K_1}\) and \({K_2}\) have been filled in between the plates of a capacitor as shown below. What will be the capacitance of the capacitor?
supporting img

1 \(\,\frac{{2{\varepsilon _o}A}}{d}\left( {\frac{{{K_1} \times {K_2}}}{{{K_1} + {K_2}}}} \right)\)
2 \(\frac{{2{\varepsilon _o}A}}{2}\left( {\frac{{{K_1} \times {K_2}}}{{{K_1} + {K_2}}}} \right)\)
3 \(\frac{{2{\varepsilon _o}A}}{d}\left( {{K_1} + {K_2}} \right)\)
4 \(\frac{{2{\varepsilon _o}A}}{2}\left( {\frac{{{K_1} + {K_2}}}{{{K_1} \times {K_2}}}} \right)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359235 A parallel plate capacitor is made of two plates of length \(l\), width \(W\) and separated by distance \(d\). A dielectric slab (dielectric constant \(K\)) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force \({\mathop{\rm F}\nolimits} = - \frac{{\partial U}}{{\partial {\rm{x}}}}\) where \(U\) is the energy of the capacitor when dielectric is inside the capacitor up to distance \(x\) (See figure). If the charge on the capacitor is \(Q\) then the force on the dielectric when it is near the edge is
supporting img

1 \(\frac{{{Q^2}W}}{{2d{{\text{l}}^2}{\varepsilon _0}}}K\)
2 \(\frac{{{Q^2}d}}{{2W{{\text{l}}^2}{\varepsilon _0}}}\left( {K - 1} \right)\)
3 \(\frac{{{Q^2}d}}{{2W{{\text{l}}^2}{\varepsilon _0}}}K\)
4 \(\frac{{{Q^2}W}}{{2d{{\text{l}}^2}{\varepsilon _0}}}\left( {K - 1} \right)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359236 Two condensers of capacities 2\(C\) and \(C\) are joined in parallel and charged upto potential \(V\). The battery is removed and the condenser of capacity \(C\) is filled completely with a medium of dielectric constant \(K\). The \(p\).\(d\). across the capacitors will now be

1 \(\frac{V}{K}\)
2 \(\frac{V}{{K + 2}}\)
3 \(\frac{{3V}}{K}\)
4 \(\frac{{3V}}{{K + 2}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359233 There is an air filled 1 \(pF\) parallel plate capacitor. When the plate separation is doubled and the space is filled with wax, the capacitance increases to 2 \(pF\). The dielectric constant of wax is

1 \(4\)
2 \(2\)
3 \(8\)
4 \(6\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359234 Two dielectric slabs of constant \({K_1}\) and \({K_2}\) have been filled in between the plates of a capacitor as shown below. What will be the capacitance of the capacitor?
supporting img

1 \(\,\frac{{2{\varepsilon _o}A}}{d}\left( {\frac{{{K_1} \times {K_2}}}{{{K_1} + {K_2}}}} \right)\)
2 \(\frac{{2{\varepsilon _o}A}}{2}\left( {\frac{{{K_1} \times {K_2}}}{{{K_1} + {K_2}}}} \right)\)
3 \(\frac{{2{\varepsilon _o}A}}{d}\left( {{K_1} + {K_2}} \right)\)
4 \(\frac{{2{\varepsilon _o}A}}{2}\left( {\frac{{{K_1} + {K_2}}}{{{K_1} \times {K_2}}}} \right)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359235 A parallel plate capacitor is made of two plates of length \(l\), width \(W\) and separated by distance \(d\). A dielectric slab (dielectric constant \(K\)) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force \({\mathop{\rm F}\nolimits} = - \frac{{\partial U}}{{\partial {\rm{x}}}}\) where \(U\) is the energy of the capacitor when dielectric is inside the capacitor up to distance \(x\) (See figure). If the charge on the capacitor is \(Q\) then the force on the dielectric when it is near the edge is
supporting img

1 \(\frac{{{Q^2}W}}{{2d{{\text{l}}^2}{\varepsilon _0}}}K\)
2 \(\frac{{{Q^2}d}}{{2W{{\text{l}}^2}{\varepsilon _0}}}\left( {K - 1} \right)\)
3 \(\frac{{{Q^2}d}}{{2W{{\text{l}}^2}{\varepsilon _0}}}K\)
4 \(\frac{{{Q^2}W}}{{2d{{\text{l}}^2}{\varepsilon _0}}}\left( {K - 1} \right)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359236 Two condensers of capacities 2\(C\) and \(C\) are joined in parallel and charged upto potential \(V\). The battery is removed and the condenser of capacity \(C\) is filled completely with a medium of dielectric constant \(K\). The \(p\).\(d\). across the capacitors will now be

1 \(\frac{V}{K}\)
2 \(\frac{V}{{K + 2}}\)
3 \(\frac{{3V}}{K}\)
4 \(\frac{{3V}}{{K + 2}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359233 There is an air filled 1 \(pF\) parallel plate capacitor. When the plate separation is doubled and the space is filled with wax, the capacitance increases to 2 \(pF\). The dielectric constant of wax is

1 \(4\)
2 \(2\)
3 \(8\)
4 \(6\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359234 Two dielectric slabs of constant \({K_1}\) and \({K_2}\) have been filled in between the plates of a capacitor as shown below. What will be the capacitance of the capacitor?
supporting img

1 \(\,\frac{{2{\varepsilon _o}A}}{d}\left( {\frac{{{K_1} \times {K_2}}}{{{K_1} + {K_2}}}} \right)\)
2 \(\frac{{2{\varepsilon _o}A}}{2}\left( {\frac{{{K_1} \times {K_2}}}{{{K_1} + {K_2}}}} \right)\)
3 \(\frac{{2{\varepsilon _o}A}}{d}\left( {{K_1} + {K_2}} \right)\)
4 \(\frac{{2{\varepsilon _o}A}}{2}\left( {\frac{{{K_1} + {K_2}}}{{{K_1} \times {K_2}}}} \right)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359235 A parallel plate capacitor is made of two plates of length \(l\), width \(W\) and separated by distance \(d\). A dielectric slab (dielectric constant \(K\)) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force \({\mathop{\rm F}\nolimits} = - \frac{{\partial U}}{{\partial {\rm{x}}}}\) where \(U\) is the energy of the capacitor when dielectric is inside the capacitor up to distance \(x\) (See figure). If the charge on the capacitor is \(Q\) then the force on the dielectric when it is near the edge is
supporting img

1 \(\frac{{{Q^2}W}}{{2d{{\text{l}}^2}{\varepsilon _0}}}K\)
2 \(\frac{{{Q^2}d}}{{2W{{\text{l}}^2}{\varepsilon _0}}}\left( {K - 1} \right)\)
3 \(\frac{{{Q^2}d}}{{2W{{\text{l}}^2}{\varepsilon _0}}}K\)
4 \(\frac{{{Q^2}W}}{{2d{{\text{l}}^2}{\varepsilon _0}}}\left( {K - 1} \right)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359236 Two condensers of capacities 2\(C\) and \(C\) are joined in parallel and charged upto potential \(V\). The battery is removed and the condenser of capacity \(C\) is filled completely with a medium of dielectric constant \(K\). The \(p\).\(d\). across the capacitors will now be

1 \(\frac{V}{K}\)
2 \(\frac{V}{{K + 2}}\)
3 \(\frac{{3V}}{K}\)
4 \(\frac{{3V}}{{K + 2}}\)
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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359233 There is an air filled 1 \(pF\) parallel plate capacitor. When the plate separation is doubled and the space is filled with wax, the capacitance increases to 2 \(pF\). The dielectric constant of wax is

1 \(4\)
2 \(2\)
3 \(8\)
4 \(6\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359234 Two dielectric slabs of constant \({K_1}\) and \({K_2}\) have been filled in between the plates of a capacitor as shown below. What will be the capacitance of the capacitor?
supporting img

1 \(\,\frac{{2{\varepsilon _o}A}}{d}\left( {\frac{{{K_1} \times {K_2}}}{{{K_1} + {K_2}}}} \right)\)
2 \(\frac{{2{\varepsilon _o}A}}{2}\left( {\frac{{{K_1} \times {K_2}}}{{{K_1} + {K_2}}}} \right)\)
3 \(\frac{{2{\varepsilon _o}A}}{d}\left( {{K_1} + {K_2}} \right)\)
4 \(\frac{{2{\varepsilon _o}A}}{2}\left( {\frac{{{K_1} + {K_2}}}{{{K_1} \times {K_2}}}} \right)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359235 A parallel plate capacitor is made of two plates of length \(l\), width \(W\) and separated by distance \(d\). A dielectric slab (dielectric constant \(K\)) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force \({\mathop{\rm F}\nolimits} = - \frac{{\partial U}}{{\partial {\rm{x}}}}\) where \(U\) is the energy of the capacitor when dielectric is inside the capacitor up to distance \(x\) (See figure). If the charge on the capacitor is \(Q\) then the force on the dielectric when it is near the edge is
supporting img

1 \(\frac{{{Q^2}W}}{{2d{{\text{l}}^2}{\varepsilon _0}}}K\)
2 \(\frac{{{Q^2}d}}{{2W{{\text{l}}^2}{\varepsilon _0}}}\left( {K - 1} \right)\)
3 \(\frac{{{Q^2}d}}{{2W{{\text{l}}^2}{\varepsilon _0}}}K\)
4 \(\frac{{{Q^2}W}}{{2d{{\text{l}}^2}{\varepsilon _0}}}\left( {K - 1} \right)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359236 Two condensers of capacities 2\(C\) and \(C\) are joined in parallel and charged upto potential \(V\). The battery is removed and the condenser of capacity \(C\) is filled completely with a medium of dielectric constant \(K\). The \(p\).\(d\). across the capacitors will now be

1 \(\frac{V}{K}\)
2 \(\frac{V}{{K + 2}}\)
3 \(\frac{{3V}}{K}\)
4 \(\frac{{3V}}{{K + 2}}\)