Question

Deuteron and alpha particles are put $1\stackrel{o}{A}$ apart in the air. The magnitude of the intensity of the electric field due to deuteron on the alpha particle is (N/C)

• A. Zero
• B. $2.88\times {10}^{-11}\frac{\text{N}}{\text{C}}$
• C. $1.44\times {10}^{-11}\frac{\text{N}}{\text{C}}$
• D. $5.76\times {10}^{-11}\frac{\text{N}}{\text{C}}$

Answer 1

The correct option is C

$1.44\times {10}^{-11}\frac{\text{N}}{\text{C}}$

Step 1: Given data

Distance between two particles:

$$\begin{gathered} r=1\stackrel{o}{A} \\ r=1\times {10}^{-10}m \end{gathered}$$

As there is an unbalanced positive charge on the deutron, thus, the charge on the deuteron is $q=+1e=1.6\times {10}^{-19}\mathrm{C}$

Step 2: Formula used

Using the formula of the electric field:

$E=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\times \frac{q}{r^2}$

Here,

q is the charge.

E is the electric field.

r is the distance.

Step 3: Calculation

$E=9\times {10}^9\times \frac{1.6\times {10}^{-19}}{{\left(1\times {10}^{-10}\right)}^2}$

$E=1.44\times {10}^{11}N/C$

Hence, the magnitude of the intensity of the electric field due to deuteron on the alpha particle is $1.44\times {10}^{11}N/C$.