Answer
Electric field due to Q1
E 1 = k Q 1 ∣ r − r ′ ∣ 3 ( r − r ′ ) = 9 × 1 0 9 × 10 ( 5 3 ) ( − 2 i − j ) = 4.02 × 1 0 10 ( − 2 i − j ) N / C E_1=\frac{kQ1}{|r-r'|^3}(r-r')
\\=\frac{9\times10^9\times10}{(\sqrt{5}^3)}(-2i-j) \\=4.02\times10^{10}(-2i-j) N/C E 1 = ∣ r − r ′ ∣ 3 k Q 1 ( r − r ′ ) = ( 5 3 ) 9 × 1 0 9 × 10 ( − 2 i − j ) = 4.02 × 1 0 10 ( − 2 i − j ) N / C
= − ( 8.04 i + 4.02 j ) × 1 0 10 N / C =-(8.04i+4.02j) \times10^{10}N/C = − ( 8.04 i + 4.02 j ) × 1 0 10 N / C
Electric field due to Q2
E 2 = k Q 2 ∣ r − r ′ ∣ 3 ( r − r ′ ) = 9 × 1 0 9 × 10 ( 2 3 ) ( − 2 i ) = 2.28 × 1 0 10 ( − i ) N / C E_2=\frac{kQ2}{|r-r'|^3}(r-r')
\\=\frac{9\times10^9\times10}{({2}^3)}(-2i) \\=2.28\times10^{10}(-i) N/C E 2 = ∣ r − r ′ ∣ 3 k Q 2 ( r − r ′ ) = ( 2 3 ) 9 × 1 0 9 × 10 ( − 2 i ) = 2.28 × 1 0 10 ( − i ) N / C
So total electric field at this point
E = E 1 + E 2 = − ( 10.32 i + 4.02 j ) × 1 0 10 N / C E=E_1+E_2\\=-(10.32i +4.02j)\times10^{10}N/C E = E 1 + E 2 = − ( 10.32 i + 4.02 j ) × 1 0 10 N / C
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