Problem 13.15
หัวข้อ: กฎของคูลอมบ์ (Coulomb’s Law)
จากรูปด้านล่าง จงหาขนาดแรงไฟฟ้าที่กระทำกับประจุ q₃? q₂ = -q q₃ = +q q₁ = +q
คำตอบ
\(|\vec{F}_{3}| = 0.737\frac{kq^{2}}{a^{2}}\;\blacksquare\)
\(|\vec{F}_{3}| = 0.737\frac{kq^{2}}{a^{2}}\;\blacksquare\)
ดูวิธีทำ
จาก \(\vec{F}_{3} = \vec{F}_{13} + \vec{F}_{23}\)
r̂r̂
\(\vec{F}_{3} =\frac{kq_{1}q_{3}}{r_{13}^{2}}_{13} +\frac{kq_{2}q_{3}}{r_{23}^{2}}_{23}\)
îĵî
\(\vec{F}_{3} =\frac{k(q)(q)}{( 2a)^{2}}(\frac{1}{2}+\frac{1}{2}) +\frac{k(-q)(q)}{a^{2}}\)
îĵî
\(\vec{F}_{3} =\frac{kq^{2}}{2 2a^{2}}+\frac{kq^{2}}{2 2a^{2}}-\frac{kq^{2}}{a^{2}}\)
îĵ
\(\vec{F}_{3} =\frac{kq^{2}}{2 2a^{2}}(-(22- 1) + )\)
\(\therefore |\vec{F}_{3}| =\frac{kq^{2}}{2 2a^{2}}\sqrt{(-(22 - 1))^{2} + 1^{2}}\)
\(|\vec{F}_{3}| = 0.737\frac{kq^{2}}{a^{2}}\;\blacksquare\)