Question

a thin uniformly - charged non - conducting arc of radius r = 10.0 cm has total charge q = 25 nc as shown. the arc subtends an angle about the point p of 40°. find the electric potential at the point p in the figure. o 1870 v o 3190 v o 2250 v o 2760 v o 1120 v

Explanation:

Step1: Recall electric - potential formula for a point - charge

The electric potential due to a point - charge $Q$ at a distance $r$ is given by $V = \frac{kQ}{r}$, where $k=\frac{1}{4\pi\epsilon_0}=9\times 10^{9}\ N\cdot m^{2}/C^{2}$. For a continuous charge distribution, we can use the same formula by considering small charge elements $dQ$. Since the charge on the arc is uniformly distributed, and the distance from every point on the arc to the point $P$ (the center of the arc) is $r = R$.

Step2: Calculate the electric potential

We know that $Q = 25\ nC=25\times 10^{-9}\ C$ and $R = 10.0\ cm = 0.1\ m$. Using the formula $V=\frac{kQ}{R}$, we substitute the values:
\[V=\frac{(9\times 10^{9}\ N\cdot m^{2}/C^{2})\times(25\times 10^{-9}\ C)}{0.1\ m}\]
\[V=\frac{9\times25}{0.1}\ V\]
\[V = 2250\ V\]

Answer:

2250 V