ABC is an equilateral triangle of side 10m and D is the mid point of BC Charges of - Physics - Electric Charges And Fields

Question

ABC is an equilateral triangle of side 10m and D is the mid
point of BC Charges of +100 , -100 and +75 micro C are placed at B
, C and D respectively Find the force on +1 micro C charge placed
at A

Vijay Yadav · Accepted Answer

Hello,

At the point A net field along x â€“ axis E x = EB cos 60Â° + EC cos 60Â° E y = ED + EB sin 60Â° â€“ EC sin 60Â° Given that a = 10m QA = + 1 ÂµC QB = 100 ÂµC QC = â€“ 100 ÂµC QD = 75 ÂµC k = 9*109 N-m2/C2 Electric fields (magnitude) are $E_{A} = K \frac{Q_{A}}{a^{2}} = 9 \times 10^{9} \times \frac{100 \times 10^{- 6}}{10^{2}} E_{A} = 9000 N/C E_{B} = K = \frac{Q_{B}}{a^{2}} = 9 \times 10^{9} \frac{100 \times 10^{- 6}}{10^{2}} E_{B} = 9000 N/C E_{D} = K \frac{Q_{D}}{(AD)^{2}} [AD = \sqrt{a^{2} - \frac{a^{2}}{4}}, AD = \sqrt{\frac{{3 a}^{2}}{4}}]$ $= 9 \times 10^{9} \frac{75 \times 10^{- 6}}{(\frac{3 \times 10^{2}}{4})} = 900 \times 10^{(9 - 6 - 2)} E_{D} = 9000 N/C$ So, Ex = 9000 cos 60Â° + 9000 cos 60Â° = $(9000 \times \frac{1}{2}) \times 2 = 9000 N/ C$ E y = 9000 + 9000 sin 60Â° â€“ 9000 sin 60Â° = 9000 N/C So net field at point A $E_{n} = \sqrt{{E_{x}}^{2} + {E_{y}}^{2}} = \sqrt{(9000)^{2} + (9000)^{2}} = 9000 \sqrt{2} E_{n} = 12727 9 N/C$ So net force acting on point charger of (+1 Âµ C) is given by F = q E n = (1 Ã— 10â€“ 6) 12727 9 F = 12 73 Ã— 10â€“ 3N

ABC is an equilateral triangle of side 10m and D is the mid point of BC . Charges of +100 , -100 and +75 micro C are placed at B , C and D respectively . Find the force on +1 micro C charge placed at A.