Deuteron and α - particle are put 1 mathringA apart in air. Magnitude of intensity of electric field due to deuteron of α - particle is

Question

Deuteron and α - particle are put 1 mathringA apart in air. Magnitude of intensity of electric field due to deuteron of α - particle is (A) zero (B)  2.88× 1011 N/C  (C)  1.44× 1011 N/C  (D)  5.76× 1011 N/C

Tardigrade Admin · Accepted Answer

E=9 × 109 × (q/n) =9 × 109 (1.6 × 10-19/(1 × 10-10)2) =1.44 × 1011 M / C

Q. Deuteron and $α$ - particle are put $1 \overset{˚}{A}$ apart in air. Magnitude of intensity of electric field due to deuteron of $α$ - particle is

zero

$2.88 \times 10^{11} N / C$

$1.44 \times 10^{11} N / C$

$5.76 \times 10^{11} N / C$

$E = 9 \times 1 0^{9} \times \frac{q}{n}$
$= 9 \times 1 0^{9} \frac{1.6 \times 1 0 ^{- 19}}{( 1 \times 1 0 ^{- 10} ) ^{2}}$
$= 1.44 \times 1 0^{11} M / C$

Q. Deuteron and α - particle are put 1A˚ apart in air. Magnitude of intensity of electric field due to deuteron of α - particle is

zero

2.88×1011N/C

1.44×1011N/C

5.76×1011N/C

E=9×109×nq​ =9×109(1×10−10)21.6×10−19​ =1.44×1011M/C

Q. Deuteron and $α$ - particle are put $1 \overset{˚}{A}$ apart in air. Magnitude of intensity of electric field due to deuteron of $α$ - particle is

$2.88 \times 10^{11} N / C$

$1.44 \times 10^{11} N / C$

$5.76 \times 10^{11} N / C$

$E = 9 \times 1 0^{9} \times \frac{q}{n}$
$= 9 \times 1 0^{9} \frac{1.6 \times 1 0 ^{- 19}}{( 1 \times 1 0 ^{- 10} ) ^{2}}$
$= 1.44 \times 1 0^{11} M / C$