Question

Need a help...bantu yaah..

Answer 1

Cukup panjang klo di jelasin
qa=50 nc=5x10-5 c

qb=32nc=32x10-6 c

Rbp=40cm=4x10-¹m

Rba=30cm=3x10-¹m

k=9x10 ^9nm²/c²

dit=ep ???????????

Rap=√30²+40² cm

=√900+1600 cm.

=√2500 cm

=50 cm =50x10-¹ m

» sudut Apb nilainya:

sin ∅=30/50 cm
=3/5
sin ∅=37°

sudut apit antara Ea dan Eb

a=180°- =180°-37°
=143°

kuat medan A(Ea)
Ea=Kqa/Rap²

Ea=(9x10^9)(5x10^-5)/(5x10-¹)²
=1.8x10^6 N/m

Kuat medan B(Eb)

Eb=Kqb/Rbp²
=(9x10^9)(32x10^-6)/(4x10-¹)²
=1.8x10^6 N/M

Nilai Ea=Eb maka:

E=√Ea²+Eb²+2EaEb cos∅

=√2E²+2E²(-4/5)
=√2E²-8/5 E²
=√2/5E²
=E√2/5

=1.8x10^6√2/√5. x√5/√5 N/C
=1.8x10^6/5 x√10 N/C
=3,6√10 x10^5 N/C

jadi kuat medan listrik di titik p sebesar
3,6√10 x10^5 N/C

Semoga bermanfaat ^_^