Problem

Two equal and opposite charges of magnitude $2 \times 10^{-7} \, \mathrm{C}$ are placed $15 \, \mathrm{cm}$ apart.

1. What is the magnitude and direction of the electric intensity ($E$) at a point midway between the charges?
2. What force would act on a proton ($q = +1.6 \times 10^{-19} \, \mathrm{C}$) placed there?

Data

• $Q = +2 \times 10^{-7} \, \mathrm{C}$
• $-Q = -2 \times 10^{-7} \, \mathrm{C}$
• Distance between charges, $r = 15 \, \mathrm{cm} = 0.15 \, \mathrm{m}$
• Midway distance: $d_1 = d_2 = d = \frac{r}{2} = 0.075 \, \mathrm{m}$
• Coulomb’s constant, $k = 9 \times 10^9 \, \mathrm{Nm^2C^{-2}}$

Prerequisite Concepts

• Electric Field Intensity:

$$E = \frac{k Q}{r^2}$$

• Force on a Charge:

$$F = qE$$

Answer

Step 1: Electric Field Intensity Due to $Q$

The electric field intensity at the midpoint due to $Q$ is:

$$E_Q = \frac{k Q}{d^2}$$

Substitute values:

$$E_Q = \frac{(9 \times 10^9)(2 \times 10^{-7})}{(0.075)^2}$$ $$E_Q = 3.2 \times 10^5 \, \mathrm{N/C}$$

Step 2: Electric Field Intensity Due to $-Q$

The electric field intensity due to $-Q$ at the midpoint is also:

$$E_{-Q} = \frac{k Q}{d^2}$$

Substitute values:

$$E_{-Q} = 3.2 \times 10^5 \, \mathrm{N/C}$$

Step 3: Net Electric Field

Since the charges are opposite, their electric fields add up at the midpoint:

$$E_{\text{net}} = E_Q + E_{-Q}$$ $$E_{\text{net}} = 3.2 \times 10^5 + 3.2 \times 10^5$$ $$E_{\text{net}} = 6.4 \times 10^5 \, \mathrm{N/C}$$

Step 4: Force on Proton

The force on a proton placed at the midpoint is:

$$F_p = q E_{\text{net}}$$

Substitute values:

$$F_p = (1.6 \times 10^{-19})(6.4 \times 10^5)$$ $$F_p = 1.024 \times 10^{-13} \, \mathrm{N}$$

Final Answer

1. Electric Intensity: $6.4 \times 10^5 \, \mathrm{N/C}$ directed from $-Q$ to $Q$.
2. Force on Proton: $1.024 \times 10^{-13} \, \mathrm{N}$ in the same direction.

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