Physics, asked by pushplatabehera25, 5 hours ago

the potential at a point is given by V=-2x^2. Find electric field at X=5m.​

Answers

Answered by xXKaminiKanyaXx
4

Answer:

Hint.

Let A=2a,B=3b,C=4cA=2a,B=3b,C=4c . We observe that A^2=4a^2,B^2=9b^2,C^2=16c^2A

2

=4a

2

,B

2

=9b

2

,C

2

=16c

2

.

Also, we can observe that the sum of the bases of three cubes equal to zero. We have a suitable identity a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2+ab+bc+ca)a

3

+b

3

+c

3

−3abc=(a+b+c)(a

2

+b

2

+c

2

+ab+bc+ca) since it is a sum of three cubes. If the sum of bases are equal to zero, one of the factors is zero, a+b+c=0a+b+c=0 . Consequently a^3+b^3+c^3=3abca

3

+b

3

+c

3

=3abc .

Solution.

Given, \dfrac{(4a^2-9b^2)^3+(9b^2-16c^2)^3+(16c^2-4a^2)^3}{(2a-3b)^3+(3b-4c)^3+(4c-2a)^3}

(2a−3b)

3

+(3b−4c)

3

+(4c−2a)

3

(4a

2

−9b

2

)

3

+(9b

2

−16c

2

)

3

+(16c

2

−4a

2

)

3

=\dfrac{(A^2-B^2)+(B^2-C^2)^3+(C^2-A^2)^3}{(A-B)^3+(B-C)^3+(C-A)^3}=

(A−B)

3

+(B−C)

3

+(C−A)

3

(A

2

−B

2

)+(B

2

−C

2

)

3

+(C

2

−A

2

)

3

=\dfrac{3(A^2-B^2)(B^2-C^2)(C^2-A^2)}{3(A-B)(B-C)(C-A)}=

3(A−B)(B−C)(C−A)

3(A

2

−B

2

)(B

2

−C

2

)(C

2

−A

2

)

=(A+B)(B+C)(C+A)=(A+B)(B+C)(C+A)

Substituting A=2a,B=3b,C=4cA=2a,B=3b,C=4c back into the equation,

=(2a+3b)(3b+4c)(4c+2a)=(2a+3b)(3b+4c)(4c+2a)

=\boxed{2(2a+3b)(3b+4c)(2c+a)}=

2(2a+3b)(3b+4c)(2c+a)

This is the required answer.

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