Question

Ten positively-charged particles are kept fixed on the x-axis at points x = 10 cm, 20 cm, 30 cm, ...., 100 cm. the first particle has a charge 1.0 × 10−8 C, the second 8 × 10−8 C, the third 27 × 10−8 C and so on. The tenth particle has a charge 1000 × 10−8 C. Find the magnitude of the electric force acting on a 1 C charge placed at the origin.

Answer 1

Answer:

495 KN

Explanation:

first force = kq1q/d1² = k(q1/d1²) = 2k(1³/0.1²)  ×10^-8

resultant force

R = sum of forces = k{ 1³/0.1²+ 2³/0.2² + 3³/0.3² +  ... + 10³/1²}  ×10^-8

= k{ 100 + 200 +300 + ...+1000}  ×10^-8

= 90 { 100 + 200 +300 + ...+1000} N

= 9000 ×55 N

= 495 KN

Answer 2

$4.95 \times 10^3 N$ is required Magnitude of Force

Explanation:

Electric force felt by 1 c due to $1 \times 10^-^8 c$

$F_1 = \frac {k \times 1 \times 10^-^8 \times 1}{(10 \times 10^-^2)^2}$

= $k \times 10^-^6$ N

electric force felt by 1 c due to $8 \times 10^-^8$c

$F_2 = \frac {k \times 8 \times 10^-^8 \times 1}{(23 \times 10^-^2)^2}$

= $\frac {k \times 8 \times 10^-^8 \times 10^2}{9}$

= $\frac {28k \times 10^-^6}{4}$

= $2k \times 10^-^6$ N

Similarly, $F_3 = \frac {k \times 27 \times 10^-^8 \times 1}{(30 \times 10^-^2)^2}$

So, $F = F_1 + F_2 + F_3 + ... F_1_0$

= $k \times 10^-^6 (1+2+3+...+10)$ N

= $k \times 10^-^6 \times \frac {10 \times 11}{2}$

= $55 \times 9 \times 10^9 \times 10^-^6$ N

= $4.95 \times 10^3 N$