165647
If the circumference of a sphere is $3 \mathrm{~m}$, the capacitance of the sphere in water is (dielectric constant of water $=80$ )
1 $4250 \mathrm{pF}$
2 $2760 \mathrm{pF}$
3 $2780 \mathrm{pF}$
4 $424 \mathrm{pF}$
Explanation:
: Given, $\mathrm{k}=80$ $\mathrm{L}=2 \pi \mathrm{r}=3 \mathrm{~m}$ $\mathrm{r}=\frac{3}{2 \pi}$ The capacitance of the sphere in water is, $\mathrm{C}=4 \pi \varepsilon_{0} k \mathrm{r}$ $\mathrm{C}=4 \pi \times 8.85 \times 10^{-12} \times 80 \times \frac{3}{2 \pi}$ $\mathrm{C}=4248 \times 10^{-12} \mathrm{~F} \sqcup 4250 \mathrm{pF}$ $\mathrm{C}=4250 \mathrm{pF}$
[AP EAMCET-25.08.2021
Capacitance
165648
A parallel plate capacitor has a capacity $C$. If a thin metal plate $(M)$ joins the two coatings $A$ and $B$ of the capacitor, its new capacitance is
1 $2 \mathrm{C}$
2 $\frac{\mathrm{C}}{2}$
3 0
4 $\infty$
Explanation:
: The thin metal plate join two parallel plate capacitor of capacitance $C$. Since, thin metal plate (M) joins the two coatings A and $\mathrm{B}$ of the capacitor, charges gets neutralized and both the plates acquire same potential. Therefore, potential difference $\mathrm{V}=0$ So, $\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}=\frac{\mathrm{Q}}{0}=\infty$ Thus, capacitance becomes infinite and can hold any amount of charge.
[AP EAMCET-05.10.2021
Capacitance
165649
In the given circuit, the charge on the capacitor is
: Current through the battery, $\mathrm{i}=\frac{\mathrm{E}}{\mathrm{r}+\mathrm{R}_{1}}$ Potential difference across $\mathrm{R}_{1}$, $\mathrm{V}=\mathrm{iR}_{1}$ $\mathrm{V}=\frac{\mathrm{ER}_{1}}{\mathrm{r}+\mathrm{R}_{1}}$ Thus, charge on capacitor $(\mathrm{Q})=\mathrm{CV}$ $\mathrm{Q}=\mathrm{C} \times \frac{\mathrm{ER}_{1}}{\mathrm{r}+\mathrm{R}_{1}}$ $\mathrm{Q}=\frac{\mathrm{CER}_{1}}{\mathrm{R}_{1}+\mathrm{r}}$
[AP EAMCET-25.04.2017
Capacitance
165651
Two tiny spheres carrying charges $1.8 \mu \mathrm{C}$ and $2.8 \mu \mathrm{C}$ are located at $40 \mathrm{~cm}$ apart. The potential at the mid-point of the line joining the two charges is
165647
If the circumference of a sphere is $3 \mathrm{~m}$, the capacitance of the sphere in water is (dielectric constant of water $=80$ )
1 $4250 \mathrm{pF}$
2 $2760 \mathrm{pF}$
3 $2780 \mathrm{pF}$
4 $424 \mathrm{pF}$
Explanation:
: Given, $\mathrm{k}=80$ $\mathrm{L}=2 \pi \mathrm{r}=3 \mathrm{~m}$ $\mathrm{r}=\frac{3}{2 \pi}$ The capacitance of the sphere in water is, $\mathrm{C}=4 \pi \varepsilon_{0} k \mathrm{r}$ $\mathrm{C}=4 \pi \times 8.85 \times 10^{-12} \times 80 \times \frac{3}{2 \pi}$ $\mathrm{C}=4248 \times 10^{-12} \mathrm{~F} \sqcup 4250 \mathrm{pF}$ $\mathrm{C}=4250 \mathrm{pF}$
[AP EAMCET-25.08.2021
Capacitance
165648
A parallel plate capacitor has a capacity $C$. If a thin metal plate $(M)$ joins the two coatings $A$ and $B$ of the capacitor, its new capacitance is
1 $2 \mathrm{C}$
2 $\frac{\mathrm{C}}{2}$
3 0
4 $\infty$
Explanation:
: The thin metal plate join two parallel plate capacitor of capacitance $C$. Since, thin metal plate (M) joins the two coatings A and $\mathrm{B}$ of the capacitor, charges gets neutralized and both the plates acquire same potential. Therefore, potential difference $\mathrm{V}=0$ So, $\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}=\frac{\mathrm{Q}}{0}=\infty$ Thus, capacitance becomes infinite and can hold any amount of charge.
[AP EAMCET-05.10.2021
Capacitance
165649
In the given circuit, the charge on the capacitor is
: Current through the battery, $\mathrm{i}=\frac{\mathrm{E}}{\mathrm{r}+\mathrm{R}_{1}}$ Potential difference across $\mathrm{R}_{1}$, $\mathrm{V}=\mathrm{iR}_{1}$ $\mathrm{V}=\frac{\mathrm{ER}_{1}}{\mathrm{r}+\mathrm{R}_{1}}$ Thus, charge on capacitor $(\mathrm{Q})=\mathrm{CV}$ $\mathrm{Q}=\mathrm{C} \times \frac{\mathrm{ER}_{1}}{\mathrm{r}+\mathrm{R}_{1}}$ $\mathrm{Q}=\frac{\mathrm{CER}_{1}}{\mathrm{R}_{1}+\mathrm{r}}$
[AP EAMCET-25.04.2017
Capacitance
165651
Two tiny spheres carrying charges $1.8 \mu \mathrm{C}$ and $2.8 \mu \mathrm{C}$ are located at $40 \mathrm{~cm}$ apart. The potential at the mid-point of the line joining the two charges is
165647
If the circumference of a sphere is $3 \mathrm{~m}$, the capacitance of the sphere in water is (dielectric constant of water $=80$ )
1 $4250 \mathrm{pF}$
2 $2760 \mathrm{pF}$
3 $2780 \mathrm{pF}$
4 $424 \mathrm{pF}$
Explanation:
: Given, $\mathrm{k}=80$ $\mathrm{L}=2 \pi \mathrm{r}=3 \mathrm{~m}$ $\mathrm{r}=\frac{3}{2 \pi}$ The capacitance of the sphere in water is, $\mathrm{C}=4 \pi \varepsilon_{0} k \mathrm{r}$ $\mathrm{C}=4 \pi \times 8.85 \times 10^{-12} \times 80 \times \frac{3}{2 \pi}$ $\mathrm{C}=4248 \times 10^{-12} \mathrm{~F} \sqcup 4250 \mathrm{pF}$ $\mathrm{C}=4250 \mathrm{pF}$
[AP EAMCET-25.08.2021
Capacitance
165648
A parallel plate capacitor has a capacity $C$. If a thin metal plate $(M)$ joins the two coatings $A$ and $B$ of the capacitor, its new capacitance is
1 $2 \mathrm{C}$
2 $\frac{\mathrm{C}}{2}$
3 0
4 $\infty$
Explanation:
: The thin metal plate join two parallel plate capacitor of capacitance $C$. Since, thin metal plate (M) joins the two coatings A and $\mathrm{B}$ of the capacitor, charges gets neutralized and both the plates acquire same potential. Therefore, potential difference $\mathrm{V}=0$ So, $\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}=\frac{\mathrm{Q}}{0}=\infty$ Thus, capacitance becomes infinite and can hold any amount of charge.
[AP EAMCET-05.10.2021
Capacitance
165649
In the given circuit, the charge on the capacitor is
: Current through the battery, $\mathrm{i}=\frac{\mathrm{E}}{\mathrm{r}+\mathrm{R}_{1}}$ Potential difference across $\mathrm{R}_{1}$, $\mathrm{V}=\mathrm{iR}_{1}$ $\mathrm{V}=\frac{\mathrm{ER}_{1}}{\mathrm{r}+\mathrm{R}_{1}}$ Thus, charge on capacitor $(\mathrm{Q})=\mathrm{CV}$ $\mathrm{Q}=\mathrm{C} \times \frac{\mathrm{ER}_{1}}{\mathrm{r}+\mathrm{R}_{1}}$ $\mathrm{Q}=\frac{\mathrm{CER}_{1}}{\mathrm{R}_{1}+\mathrm{r}}$
[AP EAMCET-25.04.2017
Capacitance
165651
Two tiny spheres carrying charges $1.8 \mu \mathrm{C}$ and $2.8 \mu \mathrm{C}$ are located at $40 \mathrm{~cm}$ apart. The potential at the mid-point of the line joining the two charges is
165647
If the circumference of a sphere is $3 \mathrm{~m}$, the capacitance of the sphere in water is (dielectric constant of water $=80$ )
1 $4250 \mathrm{pF}$
2 $2760 \mathrm{pF}$
3 $2780 \mathrm{pF}$
4 $424 \mathrm{pF}$
Explanation:
: Given, $\mathrm{k}=80$ $\mathrm{L}=2 \pi \mathrm{r}=3 \mathrm{~m}$ $\mathrm{r}=\frac{3}{2 \pi}$ The capacitance of the sphere in water is, $\mathrm{C}=4 \pi \varepsilon_{0} k \mathrm{r}$ $\mathrm{C}=4 \pi \times 8.85 \times 10^{-12} \times 80 \times \frac{3}{2 \pi}$ $\mathrm{C}=4248 \times 10^{-12} \mathrm{~F} \sqcup 4250 \mathrm{pF}$ $\mathrm{C}=4250 \mathrm{pF}$
[AP EAMCET-25.08.2021
Capacitance
165648
A parallel plate capacitor has a capacity $C$. If a thin metal plate $(M)$ joins the two coatings $A$ and $B$ of the capacitor, its new capacitance is
1 $2 \mathrm{C}$
2 $\frac{\mathrm{C}}{2}$
3 0
4 $\infty$
Explanation:
: The thin metal plate join two parallel plate capacitor of capacitance $C$. Since, thin metal plate (M) joins the two coatings A and $\mathrm{B}$ of the capacitor, charges gets neutralized and both the plates acquire same potential. Therefore, potential difference $\mathrm{V}=0$ So, $\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}=\frac{\mathrm{Q}}{0}=\infty$ Thus, capacitance becomes infinite and can hold any amount of charge.
[AP EAMCET-05.10.2021
Capacitance
165649
In the given circuit, the charge on the capacitor is
: Current through the battery, $\mathrm{i}=\frac{\mathrm{E}}{\mathrm{r}+\mathrm{R}_{1}}$ Potential difference across $\mathrm{R}_{1}$, $\mathrm{V}=\mathrm{iR}_{1}$ $\mathrm{V}=\frac{\mathrm{ER}_{1}}{\mathrm{r}+\mathrm{R}_{1}}$ Thus, charge on capacitor $(\mathrm{Q})=\mathrm{CV}$ $\mathrm{Q}=\mathrm{C} \times \frac{\mathrm{ER}_{1}}{\mathrm{r}+\mathrm{R}_{1}}$ $\mathrm{Q}=\frac{\mathrm{CER}_{1}}{\mathrm{R}_{1}+\mathrm{r}}$
[AP EAMCET-25.04.2017
Capacitance
165651
Two tiny spheres carrying charges $1.8 \mu \mathrm{C}$ and $2.8 \mu \mathrm{C}$ are located at $40 \mathrm{~cm}$ apart. The potential at the mid-point of the line joining the two charges is
