Point charges q1 and q2 of –12 nC and +12 nC, respectively, are placed 0.10 m apart (Fig.1). Compute the electric field caused by q1, the field caused by q2, and the resultant field
(i) at point a.
(ii) at point b
(iii) at point c
Answers
The resultant field at point a is 14.13 * 10⁴ N/C, at point b is 11.34 *10⁴ N/C and at point c is 1.06 *10⁴ N/C.
Given,
Charge q₁ = -12 nC = -12 *10⁻⁹ C.
Charge q₂ = +12 nC = 12 *10⁻⁹C.
To find,
Net field at a=?
Net field at b=?
Net field at c=?
Solution,
According to Coulomb's law:
Electric field (E) = F/ q
where F = 1/4πε₀ q₁q₂/r²
∴ E = 1/4πε₀ |q| /r² (1/4πε₀ =k= constant= 9*10⁹ Nm²/C²)
Now,
(i) at point a
Electric field due to q₁ = 1/k* q₁q₂ / q₁ r²
(Distance of point a from q₁ is 3 cm=r)
∴ E(q₁) = k* 12 nC / (3)².
Electric field due to q₂ E(q₂) = k* (-12nC) / ( 7)²
(Distance of point a from q₂ = 7cm)
Resultant field at point a = E(q₁) - E(q₂)
= k *12/9 - (1/k *-12/ 49)
= k (12/9 + 12/49) = k * (1.57) *10⁻⁹ / 10⁻⁴ = 14.13 * 10⁴ N/C towards q₁
(as shown in the attached diagram)
(ii) at point b
According to the direction of electric fields shown in the attached diagram :
Net Electric field at point b = E(q₂) - E(q₁)
∴ Resultant electric field = k*q (1/ (3)² - 1/(13)²)
= q/k (1/9 - 1/169) = 12*10⁻⁹ * 9 *10⁹ (0.105) =11.34 *10⁴ N/C away from q₂.
(iii) at point c
According to the direction of electric fields shown in the attached diagram:
Net Electric field at point c (Er) = E√2 (∵ E(q₂)=E(q₁) )
(Electric field is a vector quantity so vector addition rule is applied)
∴ Resultant electric field = kq/r² √2
= 9*10⁹ ₓ 12*10⁻⁹ / (12)² *10⁻⁴ √2
= 1.06 *10⁴ N/C.
Hence, the resultant field at point a will be 14.13 * 10⁴ N/C, at point b will be 11.34 *10⁴ N/C and at point c will be 1.06 *10⁴ N/C.
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