Question

A sphere of radius 20 cm,has a total charge of 3.14 micro coulomb distributed on its surface.Calculate the electric potential at points, 5 cm, 10cm, 15 cm,20 cm,..............50 cm.what will be the possible graph?

Answer 1

A sphere of radius $20\mathrm{cm}$ , has a total charge of 3.14 micro coulomb distributed on its surface. Calculate the electric potential at points, $5\mathrm{cm}$ , $10\mathrm{cm}$ , $15\mathrm{cm}$ , $20\mathrm{cm}$ , ..., $50\mathrm{cm}$ . what will be the possible graph?

Solution:

$q = 3.14 \times 10^{-6} C - \text{charge of the sphere};$

$\mathrm{R} = 20 \mathrm{~cm} = 0.2 \mathrm{~m} -$ radius of the sphere;

r - distance to the center of the sphere;

Formula for the electric potential of the sphere :

$$\varphi = \left\{ \begin{array}{l} \frac {\mathrm {q k}}{\mathrm {r}}, \text { if } \mathrm {r} \geq \mathrm {R} \\ \frac {\mathrm {q k}}{\mathrm {R}}, \text { if } \mathrm {r} < \mathrm {R} \end{array} , \text { where } \mathrm {k} = \frac {1}{4 \pi \varepsilon_ {0}} \right.$$ $$= 9 \times 1 0 ^ {9} \frac {\mathrm {N} \cdot \mathrm {m} ^ {2}}{\mathrm {C} ^ {2}}$$

Electric potential for the radius which is smaller than the sphere radius $(\mathrm{r} = 5\mathrm{cm},10\mathrm{cm},15\mathrm{cm})$

$$\varphi_ {5 \mathrm {c m}} = \varphi_ {1 0 \mathrm {c m}} = \varphi_ {1 5 \mathrm {c m}} = \varphi_ {2 0 \mathrm {c m}} = \frac {\mathrm {q k}}{\mathrm {R}} =$$ $$= \frac {3 . 1 4 \times 1 0 ^ {- 6} \mathrm {C} \cdot 9 \times 1 0 ^ {9} \frac {\mathrm {N} \cdot \mathrm {m} ^ {2}}{\mathrm {C} ^ {2}}}{0 . 2 \mathrm {m}} = 1 4 1 \times 1 0 ^ {3} \mathrm {V}$$

Electric potential for the radius which is bigger than the sphere radius $(\mathrm{r} = 20\mathrm{cm},25\mathrm{cm},30\mathrm{cm},35\mathrm{cm},40\mathrm{cm},45\mathrm{cm},50\mathrm{cm})$

$$\varphi_ {2 5 \mathrm {c m}} = \frac {\mathrm {q k}}{\mathrm {r}} = \frac {3 . 1 4 \times 1 0 ^ {- 6} \mathrm {C} \cdot 9 \times 1 0 ^ {9} \frac {\mathrm {N} \cdot \mathrm {m} ^ {2}}{\mathrm {C} ^ {2}}}{0 . 2 5 \mathrm {m}} = 1 1 3 \times 1 0 ^ {3} \mathrm {V}$$ $$\varphi_ {3 0 \mathrm {c m}} = \frac {\mathrm {q k}}{\mathrm {r}} = \frac {3 . 1 4 \times 1 0 ^ {- 6} \mathrm {C} \cdot 9 \times 1 0 ^ {9} \frac {\mathrm {N} \cdot \mathrm {m} ^ {2}}{\mathrm {C} ^ {2}}}{0 . 3 \mathrm {m}} = 9 4 \times 1 0 ^ {3} \mathrm {V}$$ $$\varphi_ {3 5 \mathrm {c m}} = \frac {\mathrm {q k}}{\mathrm {r}} = \frac {3 . 1 4 \times 1 0 ^ {- 6} \mathrm {C} \cdot 9 \times 1 0 ^ {9} \frac {\mathrm {N} \cdot \mathrm {m} ^ {2}}{\mathrm {C} ^ {2}}}{0 . 3 5 \mathrm {m}} = 8 1 \times 1 0 ^ {3} \mathrm {V}$$ $$\varphi_ {4 0 \mathrm {c m}} = \frac {\mathrm {q k}}{\mathrm {r}} = \frac {3 . 1 4 \times 1 0 ^ {- 6} \mathrm {C} \cdot 9 \times 1 0 ^ {9} \frac {\mathrm {N} \cdot \mathrm {m} ^ {2}}{\mathrm {C} ^ {2}}}{0 . 4 \mathrm {m}} = 7 1 \times 1 0 ^ {3} \mathrm {V}$$ $$\varphi_ {4 5 \mathrm {c m}} = \frac {\mathrm {q k}}{\mathrm {r}} = \frac {3 . 1 4 \times 1 0 ^ {- 6} \mathrm {C} \cdot 9 \times 1 0 ^ {9} \frac {\mathrm {N} \cdot \mathrm {m} ^ {2}}{\mathrm {C} ^ {2}}}{0 . 4 5 \mathrm {m}} = 6 3 \times 1 0 ^ {3} \mathrm {V}$$ $$\varphi_ {5 0 \mathrm {c m}} = \frac {\mathrm {q k}}{\mathrm {r}} = \frac {3 . 1 4 \times 1 0 ^ {- 6} \mathrm {C} \cdot 9 \times 1 0 ^ {9} \frac {\mathrm {N} \cdot \mathrm {m} ^ {2}}{\mathrm {C} ^ {2}}}{0 . 5 \mathrm {m}} = 5 7 \times 1 0 ^ {3} \mathrm {V}$$

Possible graph for the electric potential:

Answer: $\varphi_{5\mathrm{cm}} = \varphi_{10\mathrm{cm}} = \varphi_{15\mathrm{cm}} = \varphi_{20\mathrm{cm}} = \frac{\mathrm{qk}}{\mathrm{R}} = 141 \times 10^{3} \mathrm{~V}$

$$\varphi_ {2 5 \mathrm {c m}} = 1 1 3 \times 1 0 ^ {3} \mathrm {V}$$ $$\varphi_ {3 0 \mathrm {c m}} = 9 4 \times 1 0 ^ {3} \mathrm {V}$$ $$\varphi_ {3 5 \mathrm {c m}} = 8 1 \times 1 0 ^ {3} \mathrm {V}$$ $$\varphi_ {4 0 \mathrm {c m}} = 7 1 \times 1 0 ^ {3} \mathrm {V}$$ $$\varphi_ {4 5 \mathrm {c m}} = 6 3 \times 1 0 ^ {3} \mathrm {V}$$ $$\varphi_ {5 0 \mathrm {c m}} = 5 7 \times 1 0 ^ {3} \mathrm {V}$$