Question

The three point charges shown in the figure form an equilateral triangle with sides 4.9 cm long. What is the electric potential (relative to infinity) at the point indicated with the dot, which is equidistant from all three charges? Assume that the numbers in the figure are all accurate to two significant figures, (k = 1/4 pi epsilon_0 = 9.0 times 10^9 N middot m^2/C^2) 0.00 V 1300 V 640 V 1900 V Two tiny particles having charges q_1 = +56.0 nC and q_2 = -46.0 nC are separated by 0.500 m and held in place, as shown in the figure. A third particle, having a charge of 54.0 nC is placed at the point A. which is 0.18 m to the left of q_2. How much work is needed to move the third particle from point A to point B, which is 0.40 m to the left of q_1. All the points in the figure lie on the same line. (k = 1/4 pi epsilon_0 = 9.0 times 10^9 N middot m^2/C^2)

Answer 1

Problem # 3

DATA:

$k=\frac{1}{4\pi \epsilon _{0}}=9\times 10^{9}\, N.m^{2}/C^{2}$

Solution:

The electric potential of a puntual charge is defined as:

$V=k\frac{q}{r}\, \, \, \, \, \, \, \, \, \, \, \, \, \, (1)$

The three charges located in equilateral triangle are at the same distance from the p point. The value of is shown in the figure:

Now, applying equation (1) for the three charges we have:

$V_t=k\frac{q_{1}}{r}+k\frac{q_{2}}{r}+k\frac{q_{3}}{r}\, \, \, \, \, \, \, \, \, \, \, \, \, \,$

The point "p" is located at the same distance from the charges:

$V_t=\frac{k}{r}\left (q_{1}-q_{2}-q_{3} \right )\, \, \, \, \, \, \, \, \, \, \, \, \, \,$

But $q_{2}=q_{3}$ so:

$V_t=\frac{k}{r}\left (q_{1}-2q_{2} \right )\, \, \, \, \,\, \, \, \, \, \, \, \, \, \, \, \left ( 2 \right )$

Replacing values given in DATA we obtain:

$V_t=\frac{9\times 10^{9}\, N.m^{2}/C^{2}}{0.028\, m}\left [2\times 10^{-9}\, C-2.\left (1\times 10^{-9}\, C \right ) \right ]$

$V_t=\frac{9\times 10^{9}\, N.m^{2}/C^{2}}{0.028\, m}\left (2\times 10^{-9}\, C-2\times 10^{-9}\, C \right )$

Therefore:

${\color{Blue} V_t=0\, \, \, Volts }$

Problem # 4

DATA:

$k=9.0\times 10^{9}\, N.m^{2}/C^{2}$

Solution:

The Work is defined as:

First, we calculate the electric potential at the point "B". So:

Here, $V_{1}$ and $V_{2}$ are the potential of charges $q_{1}$ and $q_{2}$ at the point "B". Therefore:

$V_{1}=k\frac{q_{1}}{r_{1B}}$

Replacing values given in DATA:

$V_{1}=9.0\times 10^{9}\, N.m^{2}/C^{2}.\left (\frac{56.0\times 10^{-9}\, C}{0.40\, m} \right )$

Similarly, we calculate the electric potential $V_{2}$ :

$V_{2}=k\frac{q_{2}}{r_{2B}}$

Replacing values given in DATA:

$V_{2}=9.0\times 10^{9}\, N.m^{2}/C^{2}.\left (\frac{-46.0\times 10^{-9}\, C}{0.90\, m} \right )$

Therefore, replacing this results in equation (2)

Finally, replacing this result and the value of $q_{A}$ in equation (1):

$W=54.0\times 10^{-9}\, C\times 800\, V$

${\color{Blue} W=4.32\times 10^{-5}\, J}$