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Physics Ncert Solutions Class 12th Physics Class 12th Electrostatic Potential and Capacitance

2.1 Two charges 5 × 10^–8 C and –3 × 10^–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

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alok kumar singh | Contributor-Level 10

3 months ago

2.1

Let the charges be

$q_{1}$ = 5 $\times 10^{- 8}$ C and $q_{2}$ = -3 $\times 10^{- 8}$ C

Distance between the two charges, d = 16 cm = 0.16 m

Consider a point P on the line joining the two charges

R = distance of the point P from the charge $q_{1}$

Let the electrical potential (V) at point P is zero

Potential at point P caused by charges $q_{1}$ and $q_{2}$ respectively.

V = $\frac{1}{4 π ε_{0}} \times \frac{q_{1}}{r}$ $+$ $\frac{1}{4 π ε_{0}} \times \frac{q_{2}}{(d - r)}$ ………………………….(1)

Where $ε_{0}$ = permittivity of free space

For V = 0, the equation (1) becomes

$\frac{1}{4 π ε_{0}} \times \frac{q_{1}}{r} = - \frac{1}{4 π ε_{0}} \times \frac{q_{2}}{(d - r)}$ or $\frac{q_{1}}{r}$ = $- \frac{q_{2}}{(d - r)}$ or $\frac{5 \times 10^{- 8}}{r}$ = $- \frac{- 3 \times 10^{- 8}}{(d - r)}$

$\frac{5 \times 10^{- 8}}{r}$ = $\frac{3 \times 10^{- 8}}{(d - r)}$ or $\frac{r}{(d - r)}$ =&nbs

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2.1

Let the charges be

$q_{1}$ = 5 $\times 10^{- 8}$ C and $q_{2}$ = -3 $\times 10^{- 8}$ C

Distance between the two charges, d = 16 cm = 0.16 m

Consider a point P on the line joining the two charges

R = distance of the point P from the charge $q_{1}$

Let the electrical potential (V) at point P is zero

Potential at point P caused by charges $q_{1}$ and $q_{2}$ respectively.

V = $\frac{1}{4 π ε_{0}} \times \frac{q_{1}}{r}$ $+$ $\frac{1}{4 π ε_{0}} \times \frac{q_{2}}{(d - r)}$ ………………………….(1)

Where $ε_{0}$ = permittivity of free space

For V = 0, the equation (1) becomes

$\frac{5 \times 10^{- 8}}{r}$ = $\frac{3 \times 10^{- 8}}{(d - r)}$ or $\frac{r}{(d - r)}$ = $\frac{5 \times 10^{- 8}}{3 \times 10^{- 8}}$ or $\frac{r}{(d - r)}$ = $\frac{5}{3}$

3r = 5d -5r or r = (5d/8) = 0.1 m = 10 cm

Therefore the potential is zero at a distance of 10 cm from the charge $q_{1}$

Suppose the point P is outside the system of two charges, as shown in the following figure.

V = $\frac{1}{4 π ε_{0}} \times \frac{q_{1}}{s}$ $+$ $\frac{1}{4 π ε_{0}} \times \frac{q_{2}}{(s - d)}$ ……………………………(2)

For V = 0, we get

$\frac{1}{4 π ε_{0}} \times \frac{q_{1}}{s} = - \frac{1}{4 π ε_{0}} \times \frac{q_{2}}{(s - d)}$ or $\frac{q_{1}}{s} = - \frac{q_{2}}{(s - d)}$ or $\frac{5 \times 10^{- 8}}{s} = - \frac{- 3 \times 10^{- 8}}{(s - d)}$

$\frac{5}{s}$ = $\frac{3}{s - 0.16}$ or 5s – 0.8 = 3s or s = $\frac{0.8}{2}$ = 0.4 m = 40 cm

Therefore, the potential is zero at a distance 40 cm from $q_{1}$ , outside the system of charge.

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2.1 Two charges 5 × 10^–8 C and –3 × 10^–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

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2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

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2.1 Two charges 5 × 10^–8 C and –3 × 10^–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.